Question

if sin ( a + 2b ) = √3÷2 and cos ( a + 4b ) = 0 find a and b

Answer 1

Answer:

Hi...☺

Here is your answer...✌

Given that,

sin(A + 2B) = √3/2

sin(A + 2B) = sin 60°

⇒ A + 2B = 60° -----(1)

also given that,

cos(A + 4B) = 0

cos(A + 4B) = cos 90°

⇒ A + 4B = 90° -----(2)

Subtracting (1) from (2), we get

A + 4B - A - 2B = 90° - 60°

2B = 30°

⇒ B = 15°

Putting B = 15° in eq (1), we get

A + 2 (15°) = 60°

A + 30° = 60°

⇒ A = 30°

Answer 2

Answer:

hey mate

Step-by-step explanation:

here is your answer

>>>Given :-

sin (a + 2b) = √3/2

cos (a + 4b) = 0

>>>Find :-

a and b

>>>Solution :-

$\sin(a \: + \: 2b) = \frac{ \sqrt{3} }{2}$

a + 2b = 60°

$\cos(a \: + 4b) = \: 0$

a + 4b = 90

by trignometric functions √3/2 = 60° and 0 = 90°

a = 60°

b = 90°

$thanku$

#BeBranily