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Trigonometry questions for IPMAT, with solutions

10 free Trigonometry practice questions for IPMAT Indore, Rohtak and JIPMAT, 4 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Trigonometry on Learn.

1. Question 1Multiple choiceWritten by IPMATwayTrigonometric Equations and Ranges

   The least value of $2\sin^{2}\theta+3\cos^{2}\theta$ is

   1. A.$3$
   2. B.$\frac{5}{2}$
   3. C.$5$
   4. D.$2$
   Show the answer and solution

   Answer: D. $2$

   Write $\sin^{2}\theta=1-\cos^{2}\theta$: $2\sin^{2}\theta+3\cos^{2}\theta=2-2\cos^{2}\theta+3\cos^{2}\theta=2+\cos^{2}\theta$. Since $0\le\cos^{2}\theta\le1$, the expression lies in $[2,3]$. Its least value is $2$, at $\cos\theta=0$ (e.g. $\theta=90^\circ$).

2. Question 2Multiple choiceWritten by IPMATwayHeights and Distances

   The shadow of a tower is $30$ m long when the sun's altitude is $30^\circ$. The length of its shadow when the sun's altitude is $60^\circ$ is

   1. A.$10$ m
   2. B.$10\sqrt{3}$ m
   3. C.$90$ m
   4. D.$15$ m
   Show the answer and solution

   Answer: A. $10$ m

   Let the tower be $h$ m high. With the sun at $30^\circ$: $\tan30^\circ=\frac{h}{30}$, so $h=\frac{30}{\sqrt{3}}=10\sqrt{3}$ m. With the sun at $60^\circ$ the shadow is $h\cot60^\circ=10\sqrt{3}\cdot\frac{1}{\sqrt{3}}=10$ m.

3. Question 3Multiple choiceWritten by IPMATwayTrigonometry in Triangles

   If $x$, $y$, $z$ are the sides of a triangle with $x\le y\le z$ and $x^{2}+y^{2}<z^{2}$, then the triangle is

   1. A.acute-angled
   2. B.obtuse-angled
   3. C.equilateral
   4. D.right-angled
   Show the answer and solution

   Answer: B. obtuse-angled

   $z$ is the longest side, so $Z$ is the largest angle. By the cosine rule $\cos Z=\frac{x^{2}+y^{2}-z^{2}}{2xy}$. The numerator is negative because $x^{2}+y^{2}<z^{2}$, so $\cos Z<0$ and $Z>90^\circ$: the triangle is obtuse-angled.

4. Question 4Multiple choiceAsked in IPMAT Indore 2020Trigonometric Identities

   The value of $\cos^{2}\frac{\pi}{8}+\cos^{2}\frac{3\pi}{8}+\cos^{2}\frac{5\pi}{8}+\cos^{2}\frac{7\pi}{8}$ is

   1. A.$1$
   2. B.$\frac{3}{2}$
   3. C.$2$
   4. D.$\frac{9}{4}$
   Show the answer and solution

   Answer: C. $2$

   Since $\cos(\pi-x)=-\cos x$, the last two terms repeat the first two. Since $\frac{3\pi}{8}=\frac{\pi}{2}-\frac{\pi}{8}$, $\cos^{2}\frac{3\pi}{8}=\sin^{2}\frac{\pi}{8}$. Hence the sum is $2(\cos^{2}\frac{\pi}{8}+\sin^{2}\frac{\pi}{8})=2$.

5. Question 5Multiple choiceAsked in JIPMAT 2024Trigonometric Identities

   Given below are two statements:

   Statement I: $\dfrac{\cot 30^\circ+1}{\cot 30^\circ-1}=2(\cos 30^\circ+1)$

   Statement II: $2\sin 45^\circ\cos 45^\circ-\tan 45^\circ\cot 45^\circ=0$

   In the light of the above statements, choose the correct answer from the options given below:

   1. A.Both Statement I and Statement II are true
   2. B.Both Statement I and Statement II are false
   3. C.Statement I is true but Statement II is false
   4. D.Statement I is false but Statement II is true
   Show the answer and solution

   Answer: A. Both Statement I and Statement II are true

   Statement I: $\cot30^\circ=\sqrt{3}$, and $\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{(\sqrt{3}+1)^{2}}{2}=2+\sqrt{3}$, while $2(\cos30^\circ+1)=\sqrt{3}+2$. Statement II: $2\sin45^\circ\cos45^\circ=1$ and $\tan45^\circ\cot45^\circ=1$, difference $0$. Both are true.

6. Question 6Multiple choiceAsked in IPMAT Indore 2021Trigonometric Equations and Ranges

   The set of all real values of $p$ for which the equation $3\sin^{2}x+12\cos x-3=p$ has at least one solution is

   1. A.$[-12,12]$
   2. B.$[-12,9]$
   3. C.$[-15,9]$
   4. D.$[-15,12]$
   Show the answer and solution

   Answer: C. $[-15,9]$

   Replace $\sin^{2}x$ by $1-\cos^{2}x$: $3-3\cos^{2}x+12\cos x-3=12t-3t^{2}$ with $t=\cos x$. The vertex of this downward parabola is at $t=2$, outside $[-1,1]$, so on $[-1,1]$ it increases from $g(-1)=-15$ to $g(1)=9$. The equation has a solution exactly when $p\in[-15,9]$.

7. Question 7Multiple choiceWritten by IPMATwayHeights and Distances

   An aeroplane flying $3000$ m high passes vertically above another aeroplane at an instant when their angles of elevation from the same point on the ground are $60^\circ$ and $45^\circ$. The vertical distance between the two aeroplanes is

   1. A.$1000(3-\sqrt{3})$ m
   2. B.$1000\sqrt{3}$ m
   3. C.$1000(3+\sqrt{3})$ m
   4. D.$3000(\sqrt{3}-1)$ m
   Show the answer and solution

   Answer: A. $1000(3-\sqrt{3})$ m

   The point is $x=\frac{3000}{\tan60^\circ}=1000\sqrt{3}$ m from the foot of the vertical. The lower aeroplane is at height $x\tan45^\circ=1000\sqrt{3}\approx1732$ m, so the planes are $3000-1000\sqrt{3}=1000(3-\sqrt{3})\approx1268$ m apart.

8. Question 8Multiple choiceWritten by IPMATwayHeights and Distances

   From a point $10$ m above a lake, the angle of elevation of a cloud is $30^\circ$ and the angle of depression of its reflection in the lake is $60^\circ$. The height of the cloud above the lake is

   1. A.$10$ m
   2. B.$20$ m
   3. C.$30$ m
   4. D.$10\sqrt{3}$ m
   Show the answer and solution

   Answer: B. $20$ m

   The reflection is $H$ m below the surface, i.e. $H+10$ below the observer. So $H-10=\frac{x}{\sqrt{3}}$ and $H+10=x\sqrt{3}$; the ratio gives $H+10=3(H-10)$, hence $H=20$ m.

9. Question 9Multiple choiceAsked in IPMAT Indore 2022Trigonometric Equations and Ranges

   For $0<\theta<\frac{\pi}{4}$, let $a=(\sin\theta)^{\sin\theta}(\log_{2}\cos\theta)$, $b=(\cos\theta)^{\sin\theta}(\log_{2}\sin\theta)$, $c=(\sin\theta)^{\cos\theta}(\log_{2}\cos\theta)$ and $d=(\sin\theta)^{\sin\theta}(\log_{2}\sin\theta)$. Then, the median value in the sequence $a,b,c,d$ is

   1. A.$\frac{a+b}{2}$
   2. B.$\frac{a+d}{2}$
   3. C.$\frac{b+c}{2}$
   4. D.$\frac{c+d}{2}$
   Show the answer and solution

   Answer: B. $\frac{a+d}{2}$

   All four are a positive power times a negative logarithm. $a$ vs $c$: same logarithm, and $(\sin\theta)^{\sin\theta}>(\sin\theta)^{\cos\theta}$ (base below $1$, smaller exponent), so $a<c$. $d$ vs $a$: same factor $(\sin\theta)^{\sin\theta}$ and $\log_{2}\sin\theta<\log_{2}\cos\theta$, so $d<a$. $b$ vs $d$: same logarithm and $(\cos\theta)^{\sin\theta}>(\sin\theta)^{\sin\theta}$, so $b<d$. Thus $b<d<a<c$ and the median of the sequence is $\frac{a+d}{2}$.

10. Question 10Multiple choiceWritten by IPMATwayTrigonometry in Triangles

    In triangle $ABC$, $\sin A:\sin B:\sin C=3:5:7$ and the longest side is $14$ cm. The area of the triangle, in cm$^2$, is

    1. A.$15\sqrt3$
    2. B.$30$
    3. C.$15$
    4. D.$30\sqrt3$
    Show the answer and solution

    Answer: A. $15\sqrt3$

    By the sine rule the sides are in the ratio $3:5:7$; with the longest $=14$ the scale factor is $2$, so $a=6$, $b=10$, $c=14$. Cosine rule for the largest angle: $\cos C=\frac{6^2+10^2-14^2}{2\cdot6\cdot10}=\frac{-60}{120}=-\frac12$, so $C=120^\circ$ (obtuse, not right, since $36+100\ne196$). Area $=\frac12ab\sin C=\frac12\cdot6\cdot10\cdot\frac{\sqrt3}2=15\sqrt3$. Heron's formula gives the same value ($\approx25.98$).