Question

Find the acute angles A and B ,A>B .If sin (A+2B)=root3/2
and cos(A+4B)=0
​

Answer 1

Explanation:

Answer:-

Given that:-

$\sf{sin(A+2B) = \dfrac{\sqrt{3}}{2}}$

$\sf{cos(A+4B) = 0}$

We need to find the value of A and B, given that A > B.

Trigonometric Table:-

This is a formulated table which gives us values in angles or radians, the value of each angles.

We know that:-

$\sf{sin \ 60 = \dfrac{\sqrt{3}}{2}}$

And, we also know that the reverse form of sin gives us the exact value if cos also.

$\sf{cos \ 90 = 0}$

So now I'm placing the values:-

$\sf{sin(A+2B) = \dfrac{\sqrt{3}}{2}}$

$\sf{A+2B = 60}$

$\sf{A = 60 - 2B }$ Equation #1

$\sf{cos(A+4B) = 0}$

$\sf{A+4B = 90}$

Replacing #1 here,

$\sf{ 60 - 2B +4B = 90}$

$\sf{ 60 + 2B = 90}$

$\sf{ 2B = 30}$

$\sf{ B = 15}$

We got the value of B as 15°. With same, we will find A.

$\sf{A = 60 - 2B }$

$\sf{A = 60 - 30 }$

$\sf{A = 30 }$

So, value of A is 30°. This also verifies the condition given A > B.

Hence, we're done!

Answer 2

$\sf \green{Given, sin(A+2B)=3 \div 2}$

cos(A+4B)=0

A>B,

$\sf \orange{We \: know \: that, sin60 \: =3 \div 2 and cos90∘=0}$

Consider,

sin(A+2B)= 3

2

Consider,

cos(A+4B)=0 and cos 90=

=0

⟹(A+4B)=90°

---------------(ii)

Solve (i) and (ii) :

(A+2B)=60°

(A+4B)=90 °

Subtracting (i) from (ii),

2B=30°

B= 2/30°

=15°

From (ii)

(A+4B)=90°

Also, B=15 °

A=90° − 60°

A=30°

∴A=30° , B=15°

$\huge \sf \orange{Hope \: it \: helps \: you !!}$