If 2sin A = 1 = √2 cos B and π/2 <A < π,​​​​​​​ 3π/2

Question

If 2sin A = 1 = √2 cos B and \(\frac {\pi}{2}\) <A < \(\pi\), \(\frac {3\pi}{2}\)

Answer 1

Given, 2sin A = 1 

∴ sin A = 1/2 

we know that, 

cos² A = 1 – sin² A = 1 – \((\frac{1}{2})^2 = 1 -\frac{1}{4}= \frac{3}{4}\) 

∴ cos A = ± \(\frac{\sqrt3}{2}\)

Since \(\frac {\pi}{2} < A < \pi\)

A lies in the 2nd quadrant.

 ∴ cos A = -\(\frac{\sqrt3}{2}\)

     tan A = \(\frac{sin A}{cos A} = \cfrac {\frac{1}{2}}{-\frac{\sqrt3}{2}} =-\frac{1}{3}\)

     Also, \(\sqrt2 \) cos B = 1

  ∴ cos B = \(\frac{1}{\sqrt2}\)

We know that,

Sin² B = 1 - cos² B = 1 -\((\frac{1}{\sqrt2})^2 \frac {1}{2} = \frac {1}{2}\)

  ∴ Sin B = ±\(\frac{1}{\sqrt2}\)

since \(\frac{3\pi}{2} < B < 2\pi\)

B lies in the 4th quadrant,