Question

The value of $\sin {40}^{\circ }{35}^{\prime }\cos {19}^{\circ }{25}^{\prime }+\cos {40}^{\circ }{35}^{\prime }\sin {19}^{\circ }{25}^{\prime }$ is

• A. $\frac{1}{2}$
• B. $\frac{\sqrt{3}}{2}$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $\frac{\sqrt{3}}{2}$

$\sin {40}^{\circ }{35}^{\prime }\cos {19}^{\circ }{25}^{\prime }+\cos {40}^{\circ }{35}^{\prime }\sin {19}^{\circ }{25}^{\prime }$

$[\because \sin A\cos B+\cos A\sin B=\sin (A+B\left)\right]$

$=\sin ({40}^{\circ }{35}^{\prime }+{19}^{\circ }{25}^{\prime })$

$=\sin \left({59}^{\circ }{60}^{\prime }\right)$
$=\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$

Hence, correct answer is option (A).