1
BITSAT 2025
MCQ (Single Correct Answer)
+3
-1

The angles of a triangle are in the ratio $1: 2: 7$. The ratio of the greatest side to the least side is $(k+1):(k-1)$. The value of $k$ is

A

5

B

4

C

$\sqrt{5}$

D

1

2
BITSAT 2025
MCQ (Single Correct Answer)
+3
-1

Let $A, B$ and $C$ be the angles of a triangle and $\tan \frac{A}{2}=\frac{2}{5}, \tan \frac{B}{2}=\frac{3}{7}$. Then, $\tan \frac{C}{2}$ is equal to

A

$\frac{3}{5}$

B

$\frac{4}{7}$

C

5

D

1

3
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} $
4
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} $

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