Trigonometric Equations Question 190
Question: If in a triangle ABC, a, b, c and angle A is given and $ c\sin A<a<c, $ then
[UPSEAT 1999]
Options:
A) $ b_1+b_2=2c\cos A $
B) $ b_1+b_2=c\cos A $
C) $ b_1+b_2=3c\cos A $
D) $ b_1+b_2=4c\sin A $
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Answer:
Correct Answer: A
Solution:
- From cosine formula, $ \cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc} $ or $ b^{2}-(2c\cos A)b+(c^{2}-a^{2})=0, $ which is quadratic equation in b. $ \because c\sin A<a<c $
$ \therefore $ Two triangles will be obtained, but this is possible when two values of third side are also obtained. Clearly two values of side b will be b1 and $ b_2 $ . Let these are roots of above equation.
$ \therefore $ Sum of roots $ =b_1+b_2 $ = $ 2c\cos A $ .
BETA
