Question

Plz answer q7 2 part

Answer 1

Question: Find A and B if,

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

$\sf cos(2A -B) = \frac{1}{2}$

Given that,

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

$\boxed{\sf We \ know \ that, \frac{\sqrt{3}}{2} = sin \ 60^\circ}$

$\implies \sf sin(A + 2B) = sin60^\circ$

[Ratios of angles are equal, therefore angles are also equal]

$\implies \sf A + 2B = 60^\circ$ → Eq1

Also Given that,

$\implies \sf cos(2A - B) = \frac{1}{2}$

$\boxed{\sf We \ know \ that, \frac{1}{2} = cos \ 60^\circ}$

$\implies \sf cos(2A - B) = cos \ 60^\circ$

[Ratios of angles are equal, therefore angles are also equal]

$\implies \sf 2A - B = 60^\circ$ → Eq2

Multiplying Equation 2 by 2 we get,

$\implies \sf 4A - 2B = 120^\circ$ → Eq3

Now, Adding Eq1 and Eq3 we get,

A + 2B = 60°

4A - 2B = 120°

$\rule{100}{1}$

5A = 180°

$\sf A = \dfrac{180^{\circ}}{5}$

A = 36°

Substituting 'A' in Equation 1 we get,

A + 2B = 60°

36° + 2B = 60°

2B = 60° - 36°

2B = 24°

$\sf B = \dfrac{24^{\circ}}{2}$

B = 12°

Answers.

∴ ∠A = 36°

∴ ∠B = 12°