question_answer

The altitude AD of a \[\Delta \,\,ABC\] in which \[\angle A\] is obtuse, is 10 cm. If \[BD=10\text{ }cm\]and \[CD=10\sqrt{3}\,\,cm,\] then \[\angle A\] is:

A) 60°                   

B) 30°

C) 105°                 

D) None of these

Correct Answer: C

Solution :

In  \[\Delta \,\,ADB\text{ }\Rightarrow \text{ }AD=BD\] \[\angle 1\text{ }=\text{ }\angle 2\] and \[\angle ADB=\text{ }90{}^\circ \] \[\therefore \,\,\,\,\,\,\,\angle 1\text{ }=\angle 2=45{}^\circ \] In \[\Delta \,\,ADC\Rightarrow \tan \,(\angle CAD)=\frac{10\sqrt{3}}{10}\] \[\therefore \,\,\,\,\angle CAD=60{}^\circ \] \[\therefore \,\,\,\,\angle CAB=60{}^\circ +45{}^\circ =\underline{\mathbf{105{}^\circ }}\]