BITSAT 2024 | Properties of Triangles Question 1 | Mathematics

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

$ \frac{7}{9} $

$ \frac{2}{9} $

$ \frac{1}{3} $

$ \frac{2}{3} $

BITSAT 2024

MCQ (Single Correct Answer)

-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

50 m

$ 50 \sqrt{3} \mathrm{~m} $

$ 50 \sqrt{2} \mathrm{~m} $

$ 50(3-\sqrt{3}) \mathrm{m} $

BITSAT 2023

MCQ (Single Correct Answer)

-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

$$b^2-a^2=a^2+c^2$$

$$b^2, c^2, a^2$$ are in AP

$$c^2, a^2, b^2$$ are in AP

$$a^2, b^2, c^2$$ are in AP

BITSAT 2022

MCQ (Single Correct Answer)

-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

$$\alpha$$

$${\alpha \over 2}$$

$$2\alpha$$

$${\alpha \over 3}$$

Questions Asked from MCQ (Single Correct Answer)

BITSAT 2024 (2) BITSAT 2023 (1) BITSAT 2022 (3) BITSAT 2021 (1) BITSAT 2020 (1)

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