Question: In a $ \Delta ABC, $ if $ \angle A=115{}^\circ , $ $ \angle C=20{}^\circ $ and D is s point on BC such that $ AD\bot BC $ and $ BD=7cm, $ then AD is of length
Options:
A) $ 15cm $
B) $ 5cm $
C) $ 7cm $
D) $ 10cm $
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta ABC, $
$ \angle A=115{}^\circ $
and $ \angle C=20 $
$ \therefore $ $ \angle B=180{}^\circ -(115{}^\circ +20{}^\circ )=45{}^\circ $
Now, in $ \Delta ABD $
$ \frac{AD}{BD}=\tan 45{}^\circ $
$ \Rightarrow $ $ AD=BD=7cm $