Let 2 sin a + 3 cos b = 3 and 3 sin b + 2 cos a = 4 then

$a + b = (4 n + 1) \frac{π}{2}, n ϵ I$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$a + b = (2 n + 1) \frac{π}{2}, n ϵ I$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a and b can be two non right angles of a 3 - 4 - 5 triangle with a > b.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a and b can be two non right angles of a 3 - 4 - 5 triangle with b > a.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct options are
A
$a + b = (4 n + 1) \frac{π}{2}, n ϵ I$

D
a and b can be two non right angles of a 3 - 4 - 5 triangle with b > a.

Squaring and adding
sin (a + b) = 1
$a + b = (4 n + 1) \frac{π}{2}, n ϵ I a + b = \frac{π}{2}, b = \frac{π}{2} - a \Rightarrow 5 s i n a = 3 \Rightarrow c o s a = \frac{4}{5}$
$s i n b = \frac{4}{5}$ and cos $b = \frac{4}{5}$
Hence b > a

Suggest Corrections

Similar questions

2 cos A + 3 cos B + 5 cos C

= 2 sin A + 3 sin B + 5 sin C = 0

8 cos 3 A + 27 cos 3 B + 125 cos C

= k cos (A + B + C)

k =

Δ A B C,

2 sin A + 3 cos B = 3 \sqrt{2}

3 sin B + 2 cos A = 1.

C

2 (sin a + sin b) x - 2 sin (a - b) y = 3

2 (cos a + cos b) x + 2 cos (a - b) y = 5

sin 2 a + sin 2 b =

{(\frac{\sqrt{3} + 2 cos A}{1 - 2 sin A})}^{- 3} + {(\frac{1 + 2 sin A}{\sqrt{3} - 2 cos A})}^{- 3}

Δ

2 sin A + \sqrt{3} sin B = \frac{5}{2}

\sqrt{3} sin A + 2 sin B = \frac{3 \sqrt{3}}{2}

=

Join BYJU'S Learning Program