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 \[\Delta \,\,ADC\Rightarrow \tan \,(\angle CAD)=\frac{10\sqrt{3}}{10}\] \[\therefore \,\,\,\,\angle CAD=60{}^\circ \] \[\therefore \,\,\,\,\angle CAB=60{}^\circ +45{}^\circ =\underline{\mathbf{105{}^\circ }}\]You need to login to perform this action.
You will be redirected in
3 sec