1
BITSAT 2023
+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
2
BITSAT 2022
+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
$${\alpha \over 3}$$
3
BITSAT 2022
+3
-1

Given that a house forms a right angle view from a window of another house, and the angle of elevation from the base of the first house to the window is 60 degrees. If the separation between the two houses is 6 meters, calculate the height of the first house.

A
8$$\sqrt3$$ m
B
6$$\sqrt3$$ m
C
4$$\sqrt3$$ m
D
None of these
4
BITSAT 2022
+3
-1

If in a $$\Delta$$ABC, 2b2 = a2 + c2, then $$\frac{\sin 3B}{\sin B}$$ is equal to

A
$${{{c^2} - {a^2}} \over {2ca}}$$
B
$${{{c^2} - {a^2}} \over {ca}}$$
C
$${{{{({c^2} - {a^2})}^2}} \over {{{(ca)}^2}}}$$
D
$${\left[ {{{{c^2} - {a^2}} \over {2ca}}} \right]^2}$$
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