JIPMAT Trigonometry PYQs

2 solved Trigonometry previous year questions (PYQs) from JIPMAT past year papers — attempt each and check the answer.

Q1:jipmat 2025QA › TrigonometryMediumQA · MCQ

Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from two ships are 30 deg and 45 deg respectively. If the lighthouse is 100 m high, the distance between two ships is approximately:

A173 m
B200 m
C273 m
D300 m

Pick an option to attempt

The Setup: We are dealing with two right-angled triangles sharing the same vertical height (the lighthouse). The ships are on *opposite* sides, meaning the total distance between them is the sum of their individual horizontal distances to the base of the lighthouse. Step 1: Calculate distance for the first ship (

d_1

). Using the tangent ratio (opposite/adjacent):

\tan(30^\circ)=\frac{100}{d_1}

\frac{1}{\sqrt{3}}=\frac{100}{d_1} \implies d_1=100\sqrt{3}

Since

\sqrt{3} \approx 1.732

d_1=100(1.732)=173.2\text{ m}

Step 2: Calculate distance for the second ship (

d_2

\tan(45^\circ)=\frac{100}{d_2}

1=\frac{100}{d_2} \implies d_2=100\text{ m}

Step 3: Add them together for the total distance.

\text{Total Distance}=d_1+d_2=173.2+100=273.2\text{ m}

Rounding to the nearest whole number gives us

273

. Final Answer: 273

Q2:jipmat 2025QA › TrigonometryHardQA · MCQ

\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = 3

, then value of

\sin^4\theta - \cos^4\theta

A $\frac{1}{5}$
B $\frac{2}{5}$
C $\frac{3}{5}$
D $\frac{4}{5}$

Pick an option to attempt

The Setup: This looks like a trig nightmare, but we can easily finesse it by isolating a single trig ratio (like

\tan\theta

) first, then substituting it back into a simplified version of the target expression. Step 1: Cross-multiply the initial equation to find the relationship between sine and cosine.

\frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{3}{1}

\sin\theta + \cos\theta = 3(\sin\theta - \cos\theta)

\sin\theta + \cos\theta = 3\sin\theta - 3\cos\theta

Group the terms together:

4\cos\theta = 2\sin\theta

Divide by

2\cos\theta

to get tangent:

\frac{\sin\theta}{\cos\theta} = \frac{4}{2} \implies \tan\theta = 2

Step 2: Build a right-angled triangle. Since

\tan\theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{2}{1}

, we can find the hypotenuse using Pythagoras.

\text{Hypotenuse}^2 = 2^2 + 1^2 = 5 \implies \text{Hypotenuse} = \sqrt{5}

This means

\sin\theta = \frac{2}{\sqrt{5}}

and

\cos\theta = \frac{1}{\sqrt{5}}

. Step 3: Simplify the target expression using the difference of squares identity:

a^2 - b^2 = (a - b)(a + b)

\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta)

Since

\sin^2\theta + \cos^2\theta = 1

(the most basic trig fact), the expression simply becomes:

\sin^2\theta - \cos^2\theta

Step 4: Plug in our values from the triangle.

\sin^2\theta = (\frac{2}{\sqrt{5}})^2 = \frac{4}{5}

\cos^2\theta = (\frac{1}{\sqrt{5}})^2 = \frac{1}{5}

\text{Final Value} = \frac{4}{5} - \frac{1}{5} = \frac{3}{5}

Final Answer: 3/5

JIPMAT Trigonometry — PYPs

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