10th Class Mathematics Introduction to Trigonometry Question Bank Case Based (MCQs) - Introduction to Trigonometry

If \[\sin (A+B)=\frac{1}{\sqrt{2}}\]and \[\cos (A-B)=\frac{\sqrt{3}}{2},\] \[0{}^\circ <A+B\le 90{}^\circ ,\] \[A>B\]then \[\angle A\]is:

A) \[45{}^\circ \]

B) \[37.5{}^\circ \]

C) \[32.5{}^\circ \]

D) \[35{}^\circ \]

Correct Answer: B

Solution :

We have, \[\sin (A+B)=\frac{1}{\sqrt{2}}=\sin 45{}^\circ \]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,A+B=45{}^\circ \] .(1)

and \[\cos (A-B)=\frac{\sqrt{3}}{2}=\cos 30{}^\circ \]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,A-B=30{}^\circ \] ...(2)

Adding eqs. (1) and (2), we get

\[(A+B)+(A-B)=45{}^\circ +30{}^\circ \,\,\Rightarrow \,\,\,2A=75{}^\circ \,\,\Rightarrow \,\,A=37.5{}^\circ \]

So, option [b] is correct.