Question

How to calculate $\sin 37$

Answer 1

We know that ${37}^{\circ }<{90}^{\circ }$

Hence ${37}^{\circ }$ is acute.

From $△ABC,\angle C$ is acute.

$\therefore \angle B={90}^{\circ }$

$\therefore \angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-(\angle A+\angle B)$

$\Rightarrow \angle C={180}^{\circ }-({37}^{\circ }+{90}^{\circ })$

$\Rightarrow \angle C={180}^{\circ }-{127}^{\circ }={53}^{\circ }$

Let $AB=3$ units,$BC=4$ units then

By Pythagoras theorem,${AC}^2={AB}^2+{BC}^2$

${AC}^2=3^2+4^2=9+16=25$

$\Rightarrow AC=5$ units

$\therefore \sin {37}^{\circ }=\frac{AB}{AC}=\frac{3}{5}$