Triangles And Properties Of Triangle Question 35
Question: In a $ \Delta ABC $ $ \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^{2}c^{2}} $ equals
Options:
A) $ {{\cos }^{2}}A $
B) $ {{\cos }^{2}}B $
C) $ {{\sin }^{2}}A $
D) $ {{\sin }^{2}}B $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We know that, $ 2s=a+b+c $
$ \therefore \frac{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}{4b^{2}c^{2}} $ $ =\frac{2s(2s-2a)(2s-2b)(2s-2c)}{4b^{2}c^{2}} $ $ =4\frac{s(s-a)}{bc}\times \frac{(s-b)(s-c)}{bc} $ $ =4{{\cos }^{2}}\frac{A}{2}\times {{\sin }^{2}}\frac{A}{2} $ $ ={{\sin }^{2}}A $
BETA
