Triangles And Properties Of Triangle Question 16
Question: Let PQR be a triangle of area $ \Delta $ with $ a=2,b=7/2 $ and $ c=5/2 $ , where a, b, and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then $ \frac{2sinP-sin2P}{2\sin P+\sin 2P} $ equals
Options:
A) $ \frac{3}{4\Delta } $
B) $ \frac{45}{4\Delta } $
C) $ {{( \frac{3}{4\Delta } )}^{2}} $
D) $ {{( \frac{45}{4\Delta } )}^{2}} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{2\sin P-2\sin P\cos P}{2\sin P+2\sin P\cos P} $ $ =\frac{1-\cos P}{1+\cos P}=\frac{2{{\sin }^{2}}\frac{P}{2}}{2{{\cos }^{2}}\frac{P}{2}} $ $ ={{\tan }^{2}}\frac{P}{2}=\frac{(s-b)(s-c)}{s(s-a)}=\frac{{{((s-b)(s-c))}^{2}}}{{{\Delta }^{2}}} $ $ =\frac{{{( ( \frac{1}{2} )( \frac{3}{2} ) )}^{2}}}{{{\Delta }^{2}}}={{( \frac{3}{4\Delta } )}^{2}} $
BETA
