Question

ln a $\Delta ABC;A=\pi /3$ and $b:c=2:3$.lf $\tan \theta =\frac{\sqrt{3}}{5},0<\theta <\pi /2$ then

• A. $B={60}^0+\theta$
• B. $C={60}^0+\theta$
• C. $B={90}^0-\theta$
• D. $C={60}^0-\theta$

Answer 1

Answer: (B)

$C={60}^0+\theta$

$\frac{b}{c}=\frac{2}{3}$

$tan\theta =\frac{\sqrt{3}}{5}$

$tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}cot\left(\frac{A}{2}\right)$

$=\frac{\frac{b}{c}-1}{\frac{b}{c}+1}cot{30}^0$

$tan\left(\frac{B-C}{2}\right)=\left(\frac{\frac{2}{3}-1}{\frac{2}{3}+1}\right)\sqrt{3}$

$tan\left(\frac{B-C}{2}\right)=-tan\theta$

$\frac{C-B}{2}=\theta$

$C-B=2\theta --\left(1\right),C+B=\frac{2\pi }{3}--\left(2\right)$

(1) + (2)

$2C=2\theta +\frac{2\pi }{3}$

$C=\theta +\frac{\pi }{3}$

$B=\frac{\pi }{3}-\theta ......eq\left(\right(1)-(2\left)\right)$.