Question

Let PQR be a triangle of area with a=2,b=7/2 and c=5/2 ,where a,b and c are the length of the side of the triangle opposite to the angles at P,Q and R respectively.Then 2sinp-sin2p/2sinp+sin2p equals

Answer 1

Given:

Let PQR be a triangle of area with a=2,b=7/2 and c=5/2 ,where a,b and c are the length of the side of the triangle opposite to the angles at P,Q and R respectively.

To find:

Then 2sinp-sin2p/2sinp+sin2p equals

Solution:

From given, we have,

Let PQR be a triangle of the area with a=2,b=7/2 and c = 5/2, where a,b and c are the length of the side of the triangle opposite to the angles at P, Q and R respectively.

2sinp-sin2p/2sinp+sin2p

we use the formula,  

sin 2x = 2 sinx cosx

$\dfrac{2\sin p - \sin 2p}{2\sin p + \sin 2p}\\\\\\= \dfrac{2\sin p - 2\sin p\cos p}{2\sin p + 2\sin p\cos p}\\\\\\=\dfrac{2\sin p (1 - \cos p)}{2\sin p (1+\cos p)}$

$\dfrac{1 - \cos p}{1+\cos p}\\\\\\=\dfrac{2\sin ^2\frac{p}{2}}{2\cos ^2\frac{p}{2}}\\\\\\=\tan ^2 \dfrac{p}{2}$

we know that,

$\tan \dfrac{p}{2} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}}\\\\\tan ^2\dfrac{p}{2} = \dfrac{(s-b)(s-c)}{s(s-a)}$

multiply and divide by (s - b)(s - c)

$=\dfrac{[(s-b)(s-c)]^2}{s(s-a)(s-b)(s-c)}$

$=\dfrac{[(\frac{1}{2} \times \frac{3}{2})^2]}{\Delta^2}\\\\\\=(\dfrac{3}{4\Delta})^2$