Trigonometric Equations Question 90
Question: In a $ \Delta ABC $ , $ a,\ b,\ A $ are given and $ c_1,\ c_2 $ are two values of the third side c. The sum of the areas of two triangles with sides $ a,\ b,\ c_1 $ and $ a,b,\ c_2 $ is
Options:
A) $ \frac{1}{2}b^{2}\sin 2A $
B) $ \frac{1}{2}a^{2}\sin 2A $
C) $ b^{2}\sin 2A $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- We have $ \cos A=\frac{c^{2}+b^{2}-a^{2}}{2bc} $
$ \Rightarrow c^{2}-2bc,\cos A+(b^{2}-a^{2})=0 $ It is given that $ c_1 $ and $ c_2 $ are roots of this equation. Therefore $ c_1+c_2=2b\cos A $ and $ c_1c_2=b^{2}-a^{2} $
$ \Rightarrow $ $ k,(\sin C_1+\sin C_2)=2k\sin B\cos A $
$ \Rightarrow $ $ \sin C_1+\sin C_2=2\sin B\cos A $
$ \Rightarrow $ Now sum of the areas of two triangles = $ \frac{1}{2}ab\sin C_1+\frac{1}{2}ab\sin C_2 $ = $ \frac{1}{2}ab(\sin C_1+\sin C_2) $ $ =\frac{1}{2}ab(2\sin B\cos A)=ab\sin B\cos A $ = $ b(a\sin B)\cos A=b(b\sin A)\cos A=\frac{1}{2}b^{2}\sin 2A $ .
BETA
