In\[\Delta ABC,\]\[\angle A=90{}^\circ ,\]\[\angle C=55{}^\circ ,\]\[AD\bot BC.\] What is the value of\[\angle BAD\]? [SSC (CGL) 2013] |
A) \[60{}^\circ \]
B) \[45{}^\circ \]
C) \[55{}^\circ \]
D) \[35{}^\circ \]
Correct Answer: C
Solution :
\[\because \]\[AD\bot BC,\]\[\angle ADC=\angle ADB=90{}^\circ \] |
\[\therefore \]\[\angle B=180{}^\circ -\,\,(90{}^\circ -55{}^\circ )=35{}^\circ \] |
[by sum property of triangle] |
Now, in\[\Delta ADC,\] the sum of three angles of a triangle is \[180{}^\circ .\] |
\[\therefore \]\[\angle ADC+\angle ACD+\angle DAC=180{}^\circ \] |
\[\Rightarrow \]\[\angle DAC=180{}^\circ -90{}^\circ -55{}^\circ =35{}^\circ \] |
\[\angle BAD=90{}^\circ -35{}^\circ =55{}^\circ \] |
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