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CBSE CLASS X – MATHEMATICS (STANDARD) SAMPLE QUESTION PAPER
SET – 2 | ACADEMIC SESSION 2026–2027
General Instructions:
- This question paper contains 38 questions divided into 5 Sections: A, B, C, D and E.
- All Questions are compulsory.
- Section A comprises 20 Multiple Choice Questions (MCQs) of 1 mark each (Q1 to Q18 are MCQs and Q19 & Q20 are Assertion–Reason based).
- Section B comprises 5 Very Short Answer (VSA) questions of 2 marks each (Q21 to Q25).
- Section C comprises 6 Short Answer (SA) questions of 3 marks each (Q26 to Q31).
- Section D comprises 4 Long Answer (LA) questions of 5 marks each (Q32 to Q35).
- Section E comprises 3 Case-Based Integrated Units of Assessment of 4 marks each (Q36 to Q38) with sub-parts of values 1, 1 and 2 marks respectively. Internal choice is provided in the 2-mark question.
- Use of calculators is strictly prohibited. Use $\pi = 22/7$ wherever required unless stated otherwise.
SECTION A – MULTIPLE CHOICE QUESTIONS (1 Mark Each)
Q1.
If $\text{LCM}(x, 18) = 36$ and $\text{HCF}(x, 18) = 2$, then the value of $x$ is:
[1]
Q2.
A quadratic polynomial whose zeroes are $-3$ and $4$ is:
[1]
Q3.
For what value of $k$ will the system of equations $x + 2y = 3$ and $5x + ky + 7 = 0$ have no solution?
[1]
Q4.
The nature of the roots of the quadratic equation $3x^2 – 4\sqrt{3}x + 4 = 0$ is:
[1]
Q5.
The 11th term of the AP: $-3, -\frac{1}{2}, 2, \dots$ is:
[1]
Q6.
If the point $P(k, 0)$ divides the line segment joining $A(2, -2)$ and $B(-7, 4)$ in the ratio $1 : 2$, then the value of $k$ is:
[1]
Q7.
In $\triangle ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $DE \parallel BC$. If $AD = 3\text{ cm}$, $AB = 7\text{ cm}$, and $AE = 4.5\text{ cm}$, then the length of $AC$ is:
[1]
Q8.
If $\tan \theta = \frac{4}{3}$, then the value of $\frac{\sin \theta + \cos \theta}{\sin \theta – \cos \theta}$ is:
[1]
Q9.
If $\cos(\alpha + \beta) = 0$, then $\sin(\alpha – \beta)$ can be reduced to:
[1]
Q10.
If two tangents inclined at an angle of $60^\circ$ are drawn to a circle of radius $3\text{ cm}$, then the length of each tangent is:
[1]
Q11.
The area of a sector of a circle of radius $6\text{ cm}$ with a central angle of $60^\circ$ is:
[1]
Q12.
If the radius of the base of a right circular cone is halved and its height is doubled, the ratio of the volume of the new cone to that of the original cone is:
[1]
Q13.
For a given frequency distribution, if $\text{Mode} = 24$ and $\text{Mean} = 60$, then the $\text{Median}$ using the empirical relationship is:
[1]
Q14.
The upper limit of the modal class of the following frequency distribution is:
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Class} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text{Frequency} & 10 & 15 & 12 & 20 & 9 \\ \hline \end{array}$$ [1]
$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Class} & 0-5 & 5-10 & 10-15 & 15-20 & 20-25 \\ \hline \text{Frequency} & 10 & 15 & 12 & 20 & 9 \\ \hline \end{array}$$ [1]
Q15.
If the probability of winning a game is $0.07$, then the probability of losing it is:
[1]
Q16.
The coordinates of the reflection of the point $P(3, -5)$ in the $x$-axis are:
[1]
Q17.
A bag contains 5 red, 8 white, and 4 green balls. One ball is drawn at random from the bag. The probability that the ball drawn is not green is:
[1]
Q18.
If $x = 3\sec\theta$ and $y = 3\tan\theta$, then the value of $x^2 – y^2$ is:
[1]
Q19.
Assertion (A): The polynomial $p(x) = x^2 + 3x + 3$ has two distinct real zeroes.
Reason (R): A quadratic polynomial $ax^2 + bx + c$ has real zeroes if its discriminant $D = b^2 – 4ac \ge 0$. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Reason (R): A quadratic polynomial $ax^2 + bx + c$ has real zeroes if its discriminant $D = b^2 – 4ac \ge 0$. [1]
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Q20.
Assertion (A): The common difference of the AP: $18, 15\frac{1}{2}, 13, \dots$ is $-2.5$.
Reason (R): The common difference of an AP is given by $d = a_n – a_{n-1}$ for any integer $n > 1$. [1]
Reason (R): The common difference of an AP is given by $d = a_n – a_{n-1}$ for any integer $n > 1$. [1]
SECTION B – VERY SHORT ANSWER QUESTIONS (2 Marks Each)
Q21.
Prove that $3 + 2\sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.
[2]
Q22.
A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.
[2]
OR
$CD$ and $GH$ are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that $D$ and $H$ lie on sides $AB$ and $FE$ of $\triangle ABC$ and $\triangle FEG$ respectively. If $\triangle ABC \sim \triangle FEG$, show that $\frac{CD}{GH} = \frac{AC}{FG}$.
Q23.
Find the ratio in which the y-axis divides the line segment joining the points $A(5, -6)$ and $B(-1, -4)$. Also, find the coordinates of the point of division.
[2]
Q24.
If $\tan(A + B) = \sqrt{3}$ and $\tan(A – B) = \frac{1}{\sqrt{3}}$, where $0^\circ < A + B \le 90^\circ$ and $A > B$, find the values of $A$ and $B$.
[2]
OR
Prove that:
$$\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2\sec A$$
Q25.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
[2]
SECTION C – SHORT ANSWER QUESTIONS (3 Marks Each)
Q26.
If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 – p(x + 1) – c$, show that $(\alpha + 1)(\beta + 1) = 1 – c$.
[3]
Q27.
A fraction becomes $\frac{9}{11}$ if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes $\frac{5}{6}$. Find the fraction.
[3]
OR
Solve the following pair of equations for $x$ and $y$:
$$\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2$$
$$\frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1$$
Q28.
If the sum of the first $n$ terms of an AP is given by $S_n = 3n^2 + 5n$, find the common difference of the AP and its 15th term.
[3]
Q29.
Prove the trigonometric identity:
$$\frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\csc A – 1}{\csc A + 1}$$
[3]
Q30.
A chord of a circle of radius $10\text{ cm}$ subtends a right angle at the centre. Find the area of the corresponding:
(a) Minor segment
(b) Major sector
(Use $\pi = 3.14$). [3]
(a) Minor segment
(b) Major sector
(Use $\pi = 3.14$). [3]
OR
An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius $45\text{ cm}$, find the area between two consecutive ribs of the umbrella.
Q31.
A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears:
(a) A two-digit number
(b) A perfect square number
(c) A number divisible by 5 [3]
(a) A two-digit number
(b) A perfect square number
(c) A number divisible by 5 [3]
SECTION D – LONG ANSWER QUESTIONS (5 Marks Each)
Q32.
A train travels $360\text{ km}$ at a uniform speed. If the speed had been $5\text{ km/h}$ more, it would have taken 1 hour less for the same journey. Find the original speed of the train.
[5]
OR
An express train takes 1 hour less than a passenger train to travel $132\text{ km}$ between Mysore and Bangalore. If the average speed of the express train is $11\text{ km/h}$ more than that of the passenger train, find the average speeds of the two trains.
Q33.
Prove that the lengths of tangents drawn from an external point to a circle are equal.
Using the above theorem, prove the following: In $\triangle ABC$, a circle touches side $BC$ at point $P$ and touches sides $AB$ and $AC$ produced at points $Q$ and $R$ respectively. Prove that: $$AQ = \frac{1}{2}(\text{Perimeter of } \triangle ABC)$$ [5]
Using the above theorem, prove the following: In $\triangle ABC$, a circle touches side $BC$ at point $P$ and touches sides $AB$ and $AC$ produced at points $Q$ and $R$ respectively. Prove that: $$AQ = \frac{1}{2}(\text{Perimeter of } \triangle ABC)$$ [5]
Q34.
A solid iron pole consists of a cylinder of height $220\text{ cm}$ and base diameter $24\text{ cm}$, which is surmounted by another cylinder of height $60\text{ cm}$ and radius $8\text{ cm}$. Find the mass of the pole, given that $1\text{ cm}^3$ of iron has approximately $8\text{ g}$ mass. (Take $\pi = 3.14$).
[5]
OR
A juice seller serves his customers using glasses. The inner diameter of the cylindrical glass is $5\text{ cm}$, but the bottom has a hemispherical raised portion which reduces the capacity of the glass. If the height of the glass is $10\text{ cm}$, find the apparent capacity of the glass and its actual capacity. (Take $\pi = 3.14$).
Q35.
The following table gives the distribution of the lifetime of 400 neon lamps:
$$\begin{array}{|c|c||c|c|} \hline \text{Lifetime (in hours)} & \text{Number of lamps} & \text{Lifetime (in hours)} & \text{Number of lamps} \\ \hline 1500-2000 & 14 & 3500-4000 & 74 \\ 2000-2500 & 56 & 4000-4500 & 62 \\ 2500-3000 & 60 & 4500-5000 & 48 \\ 3000-3500 & 86 & & \\ \hline \end{array}$$ Find the median lifetime of a lamp. [5]
$$\begin{array}{|c|c||c|c|} \hline \text{Lifetime (in hours)} & \text{Number of lamps} & \text{Lifetime (in hours)} & \text{Number of lamps} \\ \hline 1500-2000 & 14 & 3500-4000 & 74 \\ 2000-2500 & 56 & 4000-4500 & 62 \\ 2500-3000 & 60 & 4500-5000 & 48 \\ 3000-3500 & 86 & & \\ \hline \end{array}$$ Find the median lifetime of a lamp. [5]
SECTION E – CASE-BASED INTEGRATED UNITS (4 Marks Each)
Q36. Case Study 1 (Arithmetic Progression):
[4]
(a) Find the amount paid by him in the 30th installment. [1]
(b) Find the total amount paid by him in the first 30 installments. [1]
(c) What remaining balance amount does he still have to pay after clearing the 30th installment? [2]
A person agrees to repay a loan of $₹1,18,000$ by paying installments every month. He starts with a first installment of $₹1000$ and increases the payment by $₹100$ every month continuously until the entire loan is paid off.
Based on the above situation, answer the following questions:
(a) Find the amount paid by him in the 30th installment. [1]
(b) Find the total amount paid by him in the first 30 installments. [1]
(c) What remaining balance amount does he still have to pay after clearing the 30th installment? [2]
OR
If he decides to increase the monthly increment to $₹200$ per month keeping the 1st installment at $₹1000$, find the amount he will pay in the 20th installment. [2]
Q37. Case Study 2 (Coordinate Geometry):
[4]
(b) Determine whether $\triangle ABC$ formed by the three stations is equilateral, isosceles, or scalene. [1]
(c) If a fourth tower $D$ is to be installed such that $ABCD$ forms a square, find the coordinates of tower $D$. [2]
To optimize mobile network coverage, a telecommunications engineer plots three primary radio towers on a Cartesian coordinate grid: Station $A(1, 4)$, Station $B(4, 1)$, and Station $C(7, 4)$, where 1 unit represents 1 km.
(a) Calculate the direct distance between Station $A$ and Station $C$. [1]
(b) Determine whether $\triangle ABC$ formed by the three stations is equilateral, isosceles, or scalene. [1]
(c) If a fourth tower $D$ is to be installed such that $ABCD$ forms a square, find the coordinates of tower $D$. [2]
OR
Find the coordinates of the central control hub $M$ which is equidistant from all three existing stations $A, B,$ and $C$. [2]
Q38. Case Study 3 (Some Applications of Trigonometry):
[4]
(b) Find the horizontal distance of the nearer ship from the base of the lighthouse. [1]
(c) Calculate the distance between the two patrol ships. (Take $\sqrt{3} \approx 1.732$). [2]
From the top of a $75\text{ m}$ high lighthouse above sea level, a guard observes two patrol ships approaching the lighthouse directly along the same line on the sea. The angles of depression of the two ships are observed to be $30^\circ$ and $45^\circ$.
(a) Draw a neat labelled mathematical diagram representing this scenario. [1]
(b) Find the horizontal distance of the nearer ship from the base of the lighthouse. [1]
(c) Calculate the distance between the two patrol ships. (Take $\sqrt{3} \approx 1.732$). [2]
OR
If the angle of elevation of the top of the lighthouse from a third point on the sea level is $60^\circ$, find the distance of this point from the base of the lighthouse. [2]
CBSE CLASS X MATHEMATICS (STANDARD) – SOLUTIONS & MARKING SCHEME (SET – 2)
SECTION A SOLUTIONS
Q1. (c) 4 [1 Mark]
Explanation: $\text{LCM} \times \text{HCF} = a \times b \implies 36 \times 2 = 18 \times x \implies x = \frac{72}{18} = 4$.
Explanation: $\text{LCM} \times \text{HCF} = a \times b \implies 36 \times 2 = 18 \times x \implies x = \frac{72}{18} = 4$.
Q2. (c) $\frac{x^2}{2} – \frac{x}{2} – 6$ [1 Mark]
Explanation: $\text{Sum of zeroes} = -3 + 4 = 1$; $\text{Product} = (-3)(4) = -12$. The polynomial is $k(x^2 – x – 12)$. For $k = 1/2$, $p(x) = \frac{x^2}{2} – \frac{x}{2} – 6$.
Explanation: $\text{Sum of zeroes} = -3 + 4 = 1$; $\text{Product} = (-3)(4) = -12$. The polynomial is $k(x^2 – x – 12)$. For $k = 1/2$, $p(x) = \frac{x^2}{2} – \frac{x}{2} – 6$.
Q3. (a) 10 [1 Mark]
Explanation: For no solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \implies \frac{1}{5} = \frac{2}{k} \implies k = 10$. (Also $\frac{2}{10} \ne \frac{-3}{7}$).
Explanation: For no solution: $\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2} \implies \frac{1}{5} = \frac{2}{k} \implies k = 10$. (Also $\frac{2}{10} \ne \frac{-3}{7}$).
Q4. (b) Two equal real roots [1 Mark]
Explanation: $D = b^2 – 4ac = (-4\sqrt{3})^2 – 4(3)(4) = 48 – 48 = 0$. Hence, two real and equal roots.
Explanation: $D = b^2 – 4ac = (-4\sqrt{3})^2 – 4(3)(4) = 48 – 48 = 0$. Hence, two real and equal roots.
Q5. (b) 22 [1 Mark]
Explanation: $a = -3$, $d = -\frac{1}{2} – (-3) = \frac{5}{2}$. $a_{11} = a + 10d = -3 + 10\left(\frac{5}{2}\right) = -3 + 25 = 22$.
Explanation: $a = -3$, $d = -\frac{1}{2} – (-3) = \frac{5}{2}$. $a_{11} = a + 10d = -3 + 10\left(\frac{5}{2}\right) = -3 + 25 = 22$.
Q6. (d) -1 [1 Mark]
Explanation: Using section formula for x-coordinate: $k = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$.
Explanation: Using section formula for x-coordinate: $k = \frac{1(-7) + 2(2)}{1 + 2} = \frac{-7 + 4}{3} = \frac{-3}{3} = -1$.
Q7. (c) 10.5 cm [1 Mark]
Explanation: By BPT corollary: $\frac{AD}{AB} = \frac{AE}{AC} \implies \frac{3}{7} = \frac{4.5}{AC} \implies AC = \frac{7 \times 4.5}{3} = 10.5\text{ cm}$.
Explanation: By BPT corollary: $\frac{AD}{AB} = \frac{AE}{AC} \implies \frac{3}{7} = \frac{4.5}{AC} \implies AC = \frac{7 \times 4.5}{3} = 10.5\text{ cm}$.
Q8. (a) 7 [1 Mark]
Explanation: Dividing numerator and denominator by $\cos\theta$: $\frac{\tan\theta + 1}{\tan\theta – 1} = \frac{4/3 + 1}{4/3 – 1} = \frac{7/3}{1/3} = 7$.
Explanation: Dividing numerator and denominator by $\cos\theta$: $\frac{\tan\theta + 1}{\tan\theta – 1} = \frac{4/3 + 1}{4/3 – 1} = \frac{7/3}{1/3} = 7$.
Q9. (b) $\cos 2\beta$ [1 Mark]
Explanation: $\cos(\alpha + \beta) = 0 = \cos 90^\circ \implies \alpha + \beta = 90^\circ \implies \alpha = 90^\circ – \beta$.
$\sin(\alpha – \beta) = \sin(90^\circ – \beta – \beta) = \sin(90^\circ – 2\beta) = \cos 2\beta$.
Explanation: $\cos(\alpha + \beta) = 0 = \cos 90^\circ \implies \alpha + \beta = 90^\circ \implies \alpha = 90^\circ – \beta$.
$\sin(\alpha – \beta) = \sin(90^\circ – \beta – \beta) = \sin(90^\circ – 2\beta) = \cos 2\beta$.
Q10. (d) $3\sqrt{3}\text{ cm}$ [1 Mark]
Explanation: The line joining external point to centre bisects the angle between tangents: $\theta = 30^\circ$. $\tan 30^\circ = \frac{r}{L} \implies \frac{1}{\sqrt{3}} = \frac{3}{L} \implies L = 3\sqrt{3}\text{ cm}$.
Explanation: The line joining external point to centre bisects the angle between tangents: $\theta = 30^\circ$. $\tan 30^\circ = \frac{r}{L} \implies \frac{1}{\sqrt{3}} = \frac{3}{L} \implies L = 3\sqrt{3}\text{ cm}$.
Q11. (a) $\frac{132}{7}\text{ cm}^2$ [1 Mark]
Explanation: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times (6)^2 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}\text{ cm}^2$.
Explanation: $\text{Area} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times (6)^2 = \frac{1}{6} \times \frac{22}{7} \times 36 = \frac{132}{7}\text{ cm}^2$.
Q12. (b) 1 : 2 [1 Mark]
Explanation: $V_1 = \frac{1}{3}\pi r^2 h$. $V_2 = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 (2h) = \frac{1}{3}\pi \left(\frac{r^2}{4}\right)(2h) = \frac{1}{2} V_1 \implies \frac{V_2}{V_1} = \frac{1}{2}$.
Explanation: $V_1 = \frac{1}{3}\pi r^2 h$. $V_2 = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 (2h) = \frac{1}{3}\pi \left(\frac{r^2}{4}\right)(2h) = \frac{1}{2} V_1 \implies \frac{V_2}{V_1} = \frac{1}{2}$.
Q13. (a) 48 [1 Mark]
Explanation: $\text{Mode} = 3\text{Median} – 2\text{Mean} \implies 24 = 3\text{Median} – 2(60) \implies 3\text{Median} = 144 \implies \text{Median} = 48$.
Explanation: $\text{Mode} = 3\text{Median} – 2\text{Mean} \implies 24 = 3\text{Median} – 2(60) \implies 3\text{Median} = 144 \implies \text{Median} = 48$.
Q14. (b) 20 [1 Mark]
Explanation: Maximum frequency is 20, corresponding to the modal class $15 – 20$. Its upper limit is 20.
Explanation: Maximum frequency is 20, corresponding to the modal class $15 – 20$. Its upper limit is 20.
Q15. (a) 0.93 [1 Mark]
Explanation: $P(\text{losing}) = 1 – P(\text{winning}) = 1 – 0.07 = 0.93$.
Explanation: $P(\text{losing}) = 1 – P(\text{winning}) = 1 – 0.07 = 0.93$.
Q16. (b) $(3, 5)$ [1 Mark]
Explanation: In reflection across the x-axis, the x-coordinate remains unchanged while the sign of the y-coordinate is inverted: $(x, -y) \rightarrow (3, 5)$.
Explanation: In reflection across the x-axis, the x-coordinate remains unchanged while the sign of the y-coordinate is inverted: $(x, -y) \rightarrow (3, 5)$.
Q17. (b) 13/17 [1 Mark]
Explanation: Total balls $= 5 + 8 + 4 = 17$. Favourable outcomes (not green) $= 5 + 8 = 13$. $P = \frac{13}{17}$.
Explanation: Total balls $= 5 + 8 + 4 = 17$. Favourable outcomes (not green) $= 5 + 8 = 13$. $P = \frac{13}{17}$.
Q18. (b) 9 [1 Mark]
Explanation: $x^2 – y^2 = 9\sec^2\theta – 9\tan^2\theta = 9(\sec^2\theta – \tan^2\theta) = 9(1) = 9$.
Explanation: $x^2 – y^2 = 9\sec^2\theta – 9\tan^2\theta = 9(\sec^2\theta – \tan^2\theta) = 9(1) = 9$.
Q19. (d) Assertion (A) is false but Reason (R) is true. [1 Mark]
Explanation: For $p(x)$, $D = 3^2 – 4(1)(3) = 9 – 12 = -3 < 0$, so it has no real zeroes.
Explanation: For $p(x)$, $D = 3^2 – 4(1)(3) = 9 – 12 = -3 < 0$, so it has no real zeroes.
Q20. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of (A). [1 Mark]
Explanation: $d = 15.5 – 18 = -2.5$.
Explanation: $d = 15.5 – 18 = -2.5$.
SECTION B SOLUTIONS
Q21. Proof of Irrationality: [2 Marks]
- Let $3 + 2\sqrt{5}$ be rational $= \frac{a}{b}$ where $a, b$ are integers, $b \ne 0$, and $\text{HCF}(a,b) = 1$. [0.5 Mark]
- $2\sqrt{5} = \frac{a}{b} – 3 = \frac{a – 3b}{b} \implies \sqrt{5} = \frac{a – 3b}{2b}$. [0.5 Mark]
- Since $a$ and $b$ are integers, $\frac{a – 3b}{2b}$ is a rational number, which implies $\sqrt{5}$ is rational. [0.5 Mark]
- This contradicts the fact that $\sqrt{5}$ is irrational. Hence, $3 + 2\sqrt{5}$ is irrational. [0.5 Mark]
Q22. Height of Tower: [2 Marks]
Let height of tower be $h\text{ m}$. In similar right triangles formed by sun’s elevation:
$\frac{\text{Height of Pole}}{\text{Height of Tower}} = \frac{\text{Shadow of Pole}}{\text{Shadow of Tower}} \implies \frac{6}{h} = \frac{4}{28}$ [1 Mark]
$h = \frac{6 \times 28}{4} = 6 \times 7 = \mathbf{42\text{ m}}$. [1 Mark]
OR
$\triangle ABC \sim \triangle FEG \implies \angle A = \angle F$ and $\angle ACB = \angle FGE$.
Since $CD$ and $GH$ bisect $\angle ACB$ and $\angle FGE$: $\angle ACD = \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE = \angle FGH$. [1 Mark]
In $\triangle ACD$ and $\triangle FGH$: $\angle A = \angle F$ and $\angle ACD = \angle FGH \implies \triangle ACD \sim \triangle FGH$ (by AA).
Therefore, $\mathbf{\frac{CD}{GH} = \frac{AC}{FG}}$. [1 Mark]
Let height of tower be $h\text{ m}$. In similar right triangles formed by sun’s elevation:
$\frac{\text{Height of Pole}}{\text{Height of Tower}} = \frac{\text{Shadow of Pole}}{\text{Shadow of Tower}} \implies \frac{6}{h} = \frac{4}{28}$ [1 Mark]
$h = \frac{6 \times 28}{4} = 6 \times 7 = \mathbf{42\text{ m}}$. [1 Mark]
OR
$\triangle ABC \sim \triangle FEG \implies \angle A = \angle F$ and $\angle ACB = \angle FGE$.
Since $CD$ and $GH$ bisect $\angle ACB$ and $\angle FGE$: $\angle ACD = \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE = \angle FGH$. [1 Mark]
In $\triangle ACD$ and $\triangle FGH$: $\angle A = \angle F$ and $\angle ACD = \angle FGH \implies \triangle ACD \sim \triangle FGH$ (by AA).
Therefore, $\mathbf{\frac{CD}{GH} = \frac{AC}{FG}}$. [1 Mark]
Q23. Division by y-axis: [2 Marks]
Let the y-axis divide $AB$ in the ratio $k : 1$ at point $P(0, y)$.
$x = \frac{k(-1) + 1(5)}{k + 1} = 0 \implies -k + 5 = 0 \implies \mathbf{k = 5}$. Ratio is $\mathbf{5 : 1}$. [1 Mark]
$y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 – 6}{6} = \frac{-26}{6} = -\frac{13}{3}$.
Coordinates of the point of division are $\mathbf{\left(0, -\frac{13}{3}\right)}$. [1 Mark]
Let the y-axis divide $AB$ in the ratio $k : 1$ at point $P(0, y)$.
$x = \frac{k(-1) + 1(5)}{k + 1} = 0 \implies -k + 5 = 0 \implies \mathbf{k = 5}$. Ratio is $\mathbf{5 : 1}$. [1 Mark]
$y = \frac{5(-4) + 1(-6)}{5 + 1} = \frac{-20 – 6}{6} = \frac{-26}{6} = -\frac{13}{3}$.
Coordinates of the point of division are $\mathbf{\left(0, -\frac{13}{3}\right)}$. [1 Mark]
Q24. Trigonometric Value: [2 Marks]
$\tan(A + B) = \sqrt{3} \implies A + B = 60^\circ$ …(1)
$\tan(A – B) = \frac{1}{\sqrt{3}} \implies A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 90^\circ \implies \mathbf{A = 45^\circ}$.
From (1): $B = 60^\circ – 45^\circ \implies \mathbf{B = 15^\circ}$. [1 Mark]
OR
$\text{LHS} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1 + \sin A)\cos A}$ [1 Mark]
$= \frac{(\cos^2 A + \sin^2 A) + 1 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = \mathbf{2\sec A} = \text{RHS}$. [1 Mark]
$\tan(A + B) = \sqrt{3} \implies A + B = 60^\circ$ …(1)
$\tan(A – B) = \frac{1}{\sqrt{3}} \implies A – B = 30^\circ$ …(2) [1 Mark]
Adding (1) and (2): $2A = 90^\circ \implies \mathbf{A = 45^\circ}$.
From (1): $B = 60^\circ – 45^\circ \implies \mathbf{B = 15^\circ}$. [1 Mark]
OR
$\text{LHS} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1 + \sin A)\cos A}$ [1 Mark]
$= \frac{(\cos^2 A + \sin^2 A) + 1 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2 + 2\sin A}{(1 + \sin A)\cos A} = \frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = \mathbf{2\sec A} = \text{RHS}$. [1 Mark]
Q25. Supplementary Angle Theorem: [2 Marks]
Let tangents $PA$ and $PB$ touch circle at $A$ and $B$ with centre $O$.
Radius is perpendicular to tangent: $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$. [1 Mark]
In quadrilateral $OAPB$: $\angle APB + \angle AOB + \angle OAP + \angle OBP = 360^\circ$
$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ \implies \mathbf{\angle APB + \angle AOB = 180^\circ}$ (Supplementary). [1 Mark]
Let tangents $PA$ and $PB$ touch circle at $A$ and $B$ with centre $O$.
Radius is perpendicular to tangent: $\angle OAP = 90^\circ$ and $\angle OBP = 90^\circ$. [1 Mark]
In quadrilateral $OAPB$: $\angle APB + \angle AOB + \angle OAP + \angle OBP = 360^\circ$
$\angle APB + \angle AOB + 90^\circ + 90^\circ = 360^\circ \implies \mathbf{\angle APB + \angle AOB = 180^\circ}$ (Supplementary). [1 Mark]
SECTION C SOLUTIONS
Q26. Polynomials Identity: [3 Marks]
$f(x) = x^2 – px – (p + c)$. Here $a = 1, b = -p, c = -(p + c)$. [1 Mark]
$\alpha + \beta = -\frac{b}{a} = p$ and $\alpha\beta = \frac{c}{a} = -(p + c)$. [1 Mark]
$\text{LHS} = (\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = -(p + c) + p + 1 = -p – c + p + 1 = \mathbf{1 – c} = \text{RHS}$. [1 Mark]
$f(x) = x^2 – px – (p + c)$. Here $a = 1, b = -p, c = -(p + c)$. [1 Mark]
$\alpha + \beta = -\frac{b}{a} = p$ and $\alpha\beta = \frac{c}{a} = -(p + c)$. [1 Mark]
$\text{LHS} = (\alpha + 1)(\beta + 1) = \alpha\beta + (\alpha + \beta) + 1 = -(p + c) + p + 1 = -p – c + p + 1 = \mathbf{1 – c} = \text{RHS}$. [1 Mark]
Q27. Linear Equations in Two Variables: [3 Marks]
Let fraction be $\frac{x}{y}$.
$\frac{x + 2}{y + 2} = \frac{9}{11} \implies 11x + 22 = 9y + 18 \implies 11x – 9y = -4$ …(1) [1 Mark]
$\frac{x + 3}{y + 3} = \frac{5}{6} \implies 6x + 18 = 5y + 15 \implies 6x – 5y = -3$ …(2) [1 Mark]
Solving (1) and (2): Multiplying (1) by 5 and (2) by 9 gives $x = 7, y = 9$.
The required fraction is $\mathbf{\frac{7}{9}}$. [1 Mark]
OR
Let $\frac{1}{\sqrt{x}} = u$ and $\frac{1}{\sqrt{y}} = v$. Equations become: $2u + 3v = 2$ and $4u – 9v = -1$.
Multiplying first by 3: $6u + 9v = 6$. Adding to second: $10u = 5 \implies u = \frac{1}{2} \implies \sqrt{x} = 2 \implies \mathbf{x = 4}$. [1.5 Marks]
$2\left(\frac{1}{2}\right) + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3} \implies \sqrt{y} = 3 \implies \mathbf{y = 9}$. [1.5 Marks]
Let fraction be $\frac{x}{y}$.
$\frac{x + 2}{y + 2} = \frac{9}{11} \implies 11x + 22 = 9y + 18 \implies 11x – 9y = -4$ …(1) [1 Mark]
$\frac{x + 3}{y + 3} = \frac{5}{6} \implies 6x + 18 = 5y + 15 \implies 6x – 5y = -3$ …(2) [1 Mark]
Solving (1) and (2): Multiplying (1) by 5 and (2) by 9 gives $x = 7, y = 9$.
The required fraction is $\mathbf{\frac{7}{9}}$. [1 Mark]
OR
Let $\frac{1}{\sqrt{x}} = u$ and $\frac{1}{\sqrt{y}} = v$. Equations become: $2u + 3v = 2$ and $4u – 9v = -1$.
Multiplying first by 3: $6u + 9v = 6$. Adding to second: $10u = 5 \implies u = \frac{1}{2} \implies \sqrt{x} = 2 \implies \mathbf{x = 4}$. [1.5 Marks]
$2\left(\frac{1}{2}\right) + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3} \implies \sqrt{y} = 3 \implies \mathbf{y = 9}$. [1.5 Marks]
Q28. Arithmetic Progression: [3 Marks]
$S_n = 3n^2 + 5n$.
$a_1 = S_1 = 3(1)^2 + 5(1) = 8$.
$S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 \implies a_2 = S_2 – S_1 = 22 – 8 = 14$. [1 Mark]
Common difference $d = a_2 – a_1 = 14 – 8 = \mathbf{6}$. [1 Mark]
15th term: $a_{15} = a_1 + 14d = 8 + 14(6) = 8 + 84 = \mathbf{92}$. [1 Mark]
$S_n = 3n^2 + 5n$.
$a_1 = S_1 = 3(1)^2 + 5(1) = 8$.
$S_2 = 3(2)^2 + 5(2) = 12 + 10 = 22 \implies a_2 = S_2 – S_1 = 22 – 8 = 14$. [1 Mark]
Common difference $d = a_2 – a_1 = 14 – 8 = \mathbf{6}$. [1 Mark]
15th term: $a_{15} = a_1 + 14d = 8 + 14(6) = 8 + 84 = \mathbf{92}$. [1 Mark]
Q29. Trigonometric Identity: [3 Marks]
$\text{LHS} = \frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}$ [1 Mark]
$= \frac{\cos A \left(\frac{1}{\sin A} – 1\right)}{\cos A \left(\frac{1}{\sin A} + 1\right)}$ [1 Mark]
$= \frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1} = \mathbf{\frac{\csc A – 1}{\csc A + 1}} = \text{RHS}$. Hence Proved. [1 Mark]
$\text{LHS} = \frac{\cot A – \cos A}{\cot A + \cos A} = \frac{\frac{\cos A}{\sin A} – \cos A}{\frac{\cos A}{\sin A} + \cos A}$ [1 Mark]
$= \frac{\cos A \left(\frac{1}{\sin A} – 1\right)}{\cos A \left(\frac{1}{\sin A} + 1\right)}$ [1 Mark]
$= \frac{\frac{1}{\sin A} – 1}{\frac{1}{\sin A} + 1} = \mathbf{\frac{\csc A – 1}{\csc A + 1}} = \text{RHS}$. Hence Proved. [1 Mark]
Q30. Areas Related to Circles: [3 Marks]
$r = 10\text{ cm}, \theta = 90^\circ$.
(a) $\text{Area of minor segment} = \text{Area of sector} – \text{Area of right }\triangle = \frac{90}{360}\pi r^2 – \frac{1}{2}r^2$
$= \frac{1}{4}(3.14)(100) – \frac{1}{2}(100) = 78.5 – 50 = \mathbf{28.5\text{ cm}^2}$. [1.5 Marks]
(b) $\text{Area of major sector} = \frac{360 – 90}{360}\pi r^2 = \frac{270}{360}(3.14)(100) = \frac{3}{4}(314) = \mathbf{235.5\text{ cm}^2}$. [1.5 Marks]
OR
Total ribs $= 8 \implies$ Central angle between two ribs $\theta = \frac{360^\circ}{8} = 45^\circ$.
$\text{Area} = \frac{45^\circ}{360^\circ}\pi r^2 = \frac{1}{8} \times \frac{22}{7} \times (45)^2 = \frac{1}{8} \times \frac{22}{7} \times 2025 = \mathbf{\frac{22275}{28}\text{ cm}^2} \approx 795.54\text{ cm}^2$. [3 Marks]
$r = 10\text{ cm}, \theta = 90^\circ$.
(a) $\text{Area of minor segment} = \text{Area of sector} – \text{Area of right }\triangle = \frac{90}{360}\pi r^2 – \frac{1}{2}r^2$
$= \frac{1}{4}(3.14)(100) – \frac{1}{2}(100) = 78.5 – 50 = \mathbf{28.5\text{ cm}^2}$. [1.5 Marks]
(b) $\text{Area of major sector} = \frac{360 – 90}{360}\pi r^2 = \frac{270}{360}(3.14)(100) = \frac{3}{4}(314) = \mathbf{235.5\text{ cm}^2}$. [1.5 Marks]
OR
Total ribs $= 8 \implies$ Central angle between two ribs $\theta = \frac{360^\circ}{8} = 45^\circ$.
$\text{Area} = \frac{45^\circ}{360^\circ}\pi r^2 = \frac{1}{8} \times \frac{22}{7} \times (45)^2 = \frac{1}{8} \times \frac{22}{7} \times 2025 = \mathbf{\frac{22275}{28}\text{ cm}^2} \approx 795.54\text{ cm}^2$. [3 Marks]
Q31. Probability: [3 Marks]
Total outcomes $= 90$.
Total outcomes $= 90$.
- (a) Two-digit numbers (10 to 90) $= 81$ numbers. $P = \frac{81}{90} = \mathbf{\frac{9}{10}}$. [1 Mark]
- (b) Perfect squares ($1, 4, 9, 16, 25, 36, 49, 64, 81$) $= 9$ numbers. $P = \frac{9}{90} = \mathbf{\frac{1}{10}}$. [1 Mark]
- (c) Numbers divisible by 5 ($5, 10, 15, \dots, 90$) $= 18$ numbers. $P = \frac{18}{90} = \mathbf{\frac{1}{5}}$. [1 Mark]
SECTION D SOLUTIONS
Q32. Train Speed Problem: [5 Marks]
Let original speed of train be $x\text{ km/h}$. Increased speed $= (x + 5)\text{ km/h}$.
$\frac{360}{x} – \frac{360}{x + 5} = 1$ [1.5 Marks]
$360\left[\frac{x + 5 – x}{x(x + 5)}\right] = 1 \implies \frac{1800}{x^2 + 5x} = 1 \implies x^2 + 5x – 1800 = 0$ [1.5 Marks]
$(x + 45)(x – 40) = 0 \implies x = 40\text{ or } x = -45$ (Speed cannot be negative).
The original speed of the train is $\mathbf{40\text{ km/h}}$. [2 Marks]
OR
Let speed of passenger train $= x\text{ km/h}$, Express train $= (x + 11)\text{ km/h}$.
$\frac{132}{x} – \frac{132}{x + 11} = 1 \implies 132(11) = x(x + 11) \implies x^2 + 11x – 1452 = 0$. [2 Marks]
$(x + 44)(x – 33) = 0 \implies x = 33\text{ km/h}$.
Speed of Passenger train is $\mathbf{33\text{ km/h}}$ and Express train is $\mathbf{44\text{ km/h}}$. [3 Marks]
Let original speed of train be $x\text{ km/h}$. Increased speed $= (x + 5)\text{ km/h}$.
$\frac{360}{x} – \frac{360}{x + 5} = 1$ [1.5 Marks]
$360\left[\frac{x + 5 – x}{x(x + 5)}\right] = 1 \implies \frac{1800}{x^2 + 5x} = 1 \implies x^2 + 5x – 1800 = 0$ [1.5 Marks]
$(x + 45)(x – 40) = 0 \implies x = 40\text{ or } x = -45$ (Speed cannot be negative).
The original speed of the train is $\mathbf{40\text{ km/h}}$. [2 Marks]
OR
Let speed of passenger train $= x\text{ km/h}$, Express train $= (x + 11)\text{ km/h}$.
$\frac{132}{x} – \frac{132}{x + 11} = 1 \implies 132(11) = x(x + 11) \implies x^2 + 11x – 1452 = 0$. [2 Marks]
$(x + 44)(x – 33) = 0 \implies x = 33\text{ km/h}$.
Speed of Passenger train is $\mathbf{33\text{ km/h}}$ and Express train is $\mathbf{44\text{ km/h}}$. [3 Marks]
Q33. Tangents Theorem & Application: [5 Marks]
- Theorem Proof: Tangents from external point: Statement [0.5 Mark], Diagram & Construction [1 Mark], Congruency of triangles $\triangle OAP \cong \triangle OBP$ (RHS) $\implies AP = BP$ [1.5 Marks].
- Application: Tangents from external points are equal: $AQ = AR$, $BQ = BP$, and $CP = CR$.
$\text{Perimeter of } \triangle ABC = AB + BC + CA = AB + (BP + PC) + CA$
$= (AB + BQ) + (CR + CA) = AQ + AR = AQ + AQ = 2AQ$. [1.5 Marks]
$\implies \mathbf{AQ = \frac{1}{2}(\text{Perimeter of }\triangle ABC)}$. Hence Proved. [0.5 Mark]
Q34. Solid Iron Pole: [5 Marks]
Lower cylinder: $R = 12\text{ cm}, H = 220\text{ cm} \implies V_1 = \pi R^2 H = 3.14 \times (12)^2 \times 220 = 3.14 \times 144 \times 220 = 99,475.2\text{ cm}^3$. [2 Marks]
Upper cylinder: $r = 8\text{ cm}, h = 60\text{ cm} \implies V_2 = \pi r^2 h = 3.14 \times (8)^2 \times 60 = 3.14 \times 64 \times 60 = 12,057.6\text{ cm}^3$. [1.5 Marks]
$\text{Total Volume} = 99475.2 + 12057.6 = 111,532.8\text{ cm}^3$. [0.5 Mark]
$\text{Mass} = 111532.8 \times 8\text{ g} = 892,262.4\text{ g} = \mathbf{892.26\text{ kg}}$. [1 Mark]
OR
$r = 2.5\text{ cm}, h = 10\text{ cm}$.
$\text{Apparent capacity} = \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = \mathbf{196.25\text{ cm}^3}$. [2.5 Marks]
$\text{Volume of hemisphere} = \frac{2}{3}\pi r^3 = \frac{2}{3}(3.14)(2.5)^3 = \frac{2}{3}(3.14)(15.625) \approx 32.71\text{ cm}^3$. [1.5 Marks]
$\text{Actual capacity} = 196.25 – 32.71 = \mathbf{163.54\text{ cm}^3}$. [1 Mark]
Lower cylinder: $R = 12\text{ cm}, H = 220\text{ cm} \implies V_1 = \pi R^2 H = 3.14 \times (12)^2 \times 220 = 3.14 \times 144 \times 220 = 99,475.2\text{ cm}^3$. [2 Marks]
Upper cylinder: $r = 8\text{ cm}, h = 60\text{ cm} \implies V_2 = \pi r^2 h = 3.14 \times (8)^2 \times 60 = 3.14 \times 64 \times 60 = 12,057.6\text{ cm}^3$. [1.5 Marks]
$\text{Total Volume} = 99475.2 + 12057.6 = 111,532.8\text{ cm}^3$. [0.5 Mark]
$\text{Mass} = 111532.8 \times 8\text{ g} = 892,262.4\text{ g} = \mathbf{892.26\text{ kg}}$. [1 Mark]
OR
$r = 2.5\text{ cm}, h = 10\text{ cm}$.
$\text{Apparent capacity} = \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = \mathbf{196.25\text{ cm}^3}$. [2.5 Marks]
$\text{Volume of hemisphere} = \frac{2}{3}\pi r^3 = \frac{2}{3}(3.14)(2.5)^3 = \frac{2}{3}(3.14)(15.625) \approx 32.71\text{ cm}^3$. [1.5 Marks]
$\text{Actual capacity} = 196.25 – 32.71 = \mathbf{163.54\text{ cm}^3}$. [1 Mark]
Q35. Median Lifetime: [5 Marks]
Cumulative frequencies ($cf$): $14, 70, 130, 216, 290, 352, 400$. [1.5 Marks]
$N = 400 \implies N/2 = 200$. Median class is $3000 – 3500$ ($l = 3000, cf = 130, f = 86, h = 500$). [1.5 Marks]
$\text{Median} = l + \left(\frac{N/2 – cf}{f}\right) \times h = 3000 + \left(\frac{200 – 130}{86}\right) \times 500$ [1 Mark]
$= 3000 + \frac{70 \times 500}{86} = 3000 + \frac{35000}{86} = 3000 + 406.98 = \mathbf{3406.98\text{ hours}}$. [1 Mark]
Cumulative frequencies ($cf$): $14, 70, 130, 216, 290, 352, 400$. [1.5 Marks]
$N = 400 \implies N/2 = 200$. Median class is $3000 – 3500$ ($l = 3000, cf = 130, f = 86, h = 500$). [1.5 Marks]
$\text{Median} = l + \left(\frac{N/2 – cf}{f}\right) \times h = 3000 + \left(\frac{200 – 130}{86}\right) \times 500$ [1 Mark]
$= 3000 + \frac{70 \times 500}{86} = 3000 + \frac{35000}{86} = 3000 + 406.98 = \mathbf{3406.98\text{ hours}}$. [1 Mark]
SECTION E SOLUTIONS
Q36. Case Study 1: [4 Marks]
- (a) $a = 1000, d = 100 \implies a_{30} = 1000 + 29(100) = \mathbf{₹3900}$. [1 Mark]
- (b) $S_{30} = \frac{30}{2}[2(1000) + 29(100)] = 15[2000 + 2900] = 15(4900) = \mathbf{₹73,500}$. [1 Mark]
- (c) Remaining balance $= ₹1,18,000 – ₹73,500 = \mathbf{₹44,500}$. [2 Marks]
Q37. Case Study 2: [4 Marks]
- (a) $AC = \sqrt{(7 – 1)^2 + (4 – 4)^2} = \sqrt{6^2 + 0} = \mathbf{6\text{ km}}$ (6 units). [1 Mark]
- (b) $AB = \sqrt{(4 – 1)^2 + (1 – 4)^2} = \sqrt{9 + 9} = \sqrt{18}$; $BC = \sqrt{(7 – 4)^2 + (4 – 1)^2} = \sqrt{9 + 9} = \sqrt{18}$. Since $AB = BC \ne AC$, $\triangle ABC$ is an isosceles right-angled triangle ($AB^2 + BC^2 = AC^2$). [1 Mark]
- (c) Midpoint of diagonal $AC = \left(\frac{1 + 7}{2}, \frac{4 + 4}{2}\right) = (4, 4)$.
For square $ABCD$, midpoint of $BD$ must be $(4, 4) \implies \left(\frac{4 + x_D}{2}, \frac{1 + y_D}{2}\right) = (4, 4) \implies \mathbf{D(4, 7)}$. [2 Marks]
Q38. Case Study 3: [4 Marks]
- (a) Labelled diagram showing lighthouse $AB = 75\text{ m}$, ship 1 at $C$ with angle $45^\circ$, ship 2 at $D$ with angle $30^\circ$. [1 Mark]
- (b) In $\triangle ABC$: $\tan 45^\circ = \frac{75}{BC} \implies 1 = \frac{75}{BC} \implies BC = \mathbf{75\text{ m}}$. [1 Mark]
- (c) In $\triangle ABD$: $\tan 30^\circ = \frac{75}{BD} \implies \frac{1}{\sqrt{3}} = \frac{75}{BD} \implies BD = 75\sqrt{3}\text{ m}$.
Distance between ships $= BD – BC = 75\sqrt{3} – 75 = 75(\sqrt{3} – 1) = 75(1.732 – 1) = 75(0.732) = \mathbf{54.9\text{ m}}$. [2 Marks]