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 [SSC (CPO) 2013] |
A) \[15\,\,cm\]
B) \[5\,\,cm\]
C) \[7\,\,cm\]
D) \[10\,\,cm\]
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 \[\Delta ABD\] |
\[\frac{AD}{BD}=\tan 45{}^\circ \] |
\[\Rightarrow \] \[AD=BD=7\,\,cm\] |
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