1
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
Let $ A, B $ and $ C $ are the angles of a triangle and $ \tan \frac{A}{2}=\frac{1}{3}, \tan \frac{B}{2}=\frac{2}{3} $. Then, $ \tan \frac{C}{2} $ is equal to
A
$ \frac{7}{9} $
B
$ \frac{2}{9} $
C
$ \frac{1}{3} $
D
$ \frac{2}{3} $
2
BITSAT 2024
MCQ (Single Correct Answer)
+3
-1
$ A B C $ is a triangular park with $ A B=A C=100 \mathrm{~m} $. A TV tower stands at the mid-point of $ B C $. The angles of elevation of the top of the tower at $ A $, $ B, C $ are $ 45^{\circ}, 60^{\circ}, 60^{\circ} $ respectively. The height of the tower is
A
50 m
B
$ 50 \sqrt{3} \mathrm{~m} $
C
$ 50 \sqrt{2} \mathrm{~m} $
D
$ 50(3-\sqrt{3}) \mathrm{m} $
3
BITSAT 2023
MCQ (Single Correct Answer)
+3
-1

Let $$\frac{\sin A}{\sin B}=\frac{\sin (A-C)}{\sin (C-B)}$$, where $$A, B$$ and $$C$$ are angles of a $$\triangle A B C$$. If the lengths of the sides opposite these angles are $$a, b$$ and $$c$$ respectively, then

A
$$b^2-a^2=a^2+c^2$$
B
$$b^2, c^2, a^2$$ are in AP
C
$$c^2, a^2, b^2$$ are in AP
D
$$a^2, b^2, c^2$$ are in AP
4
BITSAT 2022
MCQ (Single Correct Answer)
+3
-1

Let $$\alpha$$ be the solution of $${16^{{{\sin }^2}\theta }} + {16^{{{\cos }^2}\theta }} = 10$$ in $$\left( {0,{\pi \over 4}} \right)$$. If the shadow of a vertical pole is $${1 \over {\sqrt 3 }}$$ of its height, then the altitude of the sun is

A
$$\alpha$$
B
$${\alpha \over 2}$$
C
$$2\alpha$$
D
$${\alpha \over 3}$$
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