Triangles And Properties Of Triangle Question 38
Question: In a triangle $ ABC,2a^{2}+4b^{2}+c^{2}=4ab+2ac, $ then $ \cos B $ is equal to
Options:
A) 0
B) $ \frac{1}{8} $
C) $ \frac{3}{8} $
D) $ \frac{7}{8} $
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Answer:
Correct Answer: D
Solution:
[d] $ a^{2}+a^{2}+4b^{2}-4ab=2ac-c^{2} $
$ \Rightarrow {{(a-2b)}^{2}}+{{(a-c)}^{2}}=0 $ which is possible only when: $ a-2b=0anda-c $ $ =0 $ or $ \frac{a}{1}=\frac{b}{1/2}=\frac{c}{1}=\lambda $ (say)
$ \therefore a=\lambda ,b=\lambda /2,c=\lambda \therefore \cos B=\frac{a^{2}+c^{2}-b^{2}}{2ac} $ $ =\frac{{{\lambda }^{2}}+{{\lambda }^{2}}-\frac{{{\lambda }^{2}}}{4}}{2{{\lambda }^{2}}}=1-\frac{1}{8}=\frac{7}{8} $
BETA
