Question

Find acute angles A and B, if $sin(A+B)=\frac{√3}{2}$ and cos (A+4B)=0, A>B.

Answer 1

SOLUTION:

Given : sin (A+ 2B) = √3/2 and cos (A + 4B) = 0

LHS  = sin(A + 2B) = √3/2

sin(A + 2B) = sin 60°

[sin 60° = √3/2]

A + 2B = 60° ………………(1)

RHS = Cos(A + 4B) = 0

Cos(A + 4B) = Cos 90°

A + 4B = 90°……………….( 2)

On subtracting eq 1 & 2

A + 2B = 60°

A + 4B = 90°

(-)  (-)    (-)

--------------------

-2B = - 30°

2B = 30°

B = 30°/2

B = 15°

On putting the value of B in eq 2

A + 4B = 90°

A + 4×15° = 90°

A + 60° = 90°

A = 90° – 60°

A = 30°

Hence, the value of A = 30∘ and B = 15∘

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\mathbb{\underline{\green{Points To Remember: }}}$

(a)• Sin 60°= √3/2

(b)• Cos 90° = 0

$\mathcal{\underline{\blue{SOLUTION: }}}$

Given,
Sin (A+B) = √3/2

From (a)

Sin(A+B) = Sin 60°

=> A + B = 60°
------------------------------Eq.(1)

Now,
Cos(A+4B) = 0

From (b)

Cos(A+4B) = Cos 90°

=> A + 4B = 90°
------------------------------Eq.(2)

Do Eq.(2) - Eq.(1):

• A + 4B = 90°
• A + B = 60°
---------------------
-……-……=-……

=> 3B = 30°
=> B = 10°

From (1) :
A + B = 60
=> A = 60 - B = 60 - 10 = 50°

•°• A = 50° & B = 10° are the Required Angles

$\mathcal{\huge{\pink{Hope It Helps}}}$