Trigonometric Equations Question 276
Question: In any triangle $ AB=2,BC=4,CA=3 $ and D is mid point of BC, then
[Roorkee 1995]
Options:
A) $ \cos B=\frac{11}{6} $
B) $ \cos B=\frac{7}{8} $
C) $ AD=2.4 $
D) $ AD^{2}=2.5 $
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Answer:
Correct Answer: D
Solution:
- $ \cos B=\frac{2^{2}+4^{2}-3^{2}}{2\times 2\times 4}=\frac{11}{16} $ \ $ \frac{11}{16}=\frac{2^{2}+2^{2}-AD^{2}}{2\times 2\times 2}\Rightarrow AD^{2}=2.5 $ .
BETA
