GATE 2024 EE GA Q.10: solution and answer

GATE 2024 · EEGA Q.10 · 2 marks · MCQ

A surveyor has to measure the horizontal distance from her position to a distant reference point C. Using her position as the center, a 200 m horizontal line segment is drawn with the two endpoints A and B. Points A, B, and C are not collinear. Each of the angles $∠ C A B$ and $∠ C B A$ are measured as $87. 8^{\circ}$ . The distance (in m) of the reference point C from her position is nearest to

(A)
2603
(B)
2606
(C)
2306
(D)
2063

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Answer and solution

ANSWER

The distance from the surveyor to C is $100 tan 87. 8^{\circ} \approx 2603 m$ , so the answer is option (A).

Official answer: ASource: Official answer key, GATE 2024 EE (IISc Bengaluru)

Let the surveyor stand at M. M is the centre of the 200 m segment AB, so $A M = M B = 100 m$ .
The base angles are equal, $∠ C A B = ∠ C B A = 87. 8^{\circ}$ , so triangle ABC is isosceles with $C A = C B$ .
In an isosceles triangle the median from the apex to the base is also the altitude. So $C M ⊥ A B$ , and triangle CMA has its right angle at M.
Right-triangle trigonometry in triangle CMA gives $tan (∠ C A M) = \frac{C M}{A M}$ , so $C M = 100 tan 87. 8^{\circ} m$ .
Complementary angle: $tan 87. 8^{\circ} = cot 2. 2^{\circ} = \frac{1}{tan 2. 2 ^{\circ}}$ .
Small angle: $2. 2^{\circ} = 2.2 \times \frac{π}{180} = 0.038397 rad$ . Then $tan x \approx x + \frac{x ^{3}}{3} = 0.038397 + 0.0000189 = 0.038416$ . The next term of the series, $\frac{2 x ^{5}}{15} \approx 1.2 \times 1 0^{- 8}$ , is too small to matter.
$C M = \frac{100}{0.038416} \approx 2603.1 m$
The nearest option is 2603, option (A).

KEY IDEA

Equal base angles make triangle ABC isosceles. The surveyor stands at the midpoint M of AB, so CM is both the median and the altitude. That gives a right triangle with a 100 m leg next to the 87.8° angle, so $C M = 100 tan 87. 8^{\circ} = 100/ tan 2. 2^{\circ}$ .

WATCH OUT FOR

The usual slip is working out the distance from an endpoint instead of from the surveyor. The slant side CA is the hypotenuse: $C A = 100/ cos 87. 8^{\circ} = 100/ sin 2. 2^{\circ} \approx 2605.0 m$ , which rounds to the distractor 2606 (B). The required distance is the altitude CM, the leg opposite the 87.8° angle. The approximation needs care as well. Taking $tan 2. 2^{\circ} \approx 2. 2^{\circ}$ in radians gives about 2604.4, which is only just closer to 2603 than to 2606. Keep the $x^{3} /3$ term, or use a table value $tan 2. 2^{\circ} \approx 0.03842$ , to get a clear answer of 2603.

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GATE 2024 EE GA Q.10: solution and answer

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