Question

Find the value of $\sin {37}^o,\sin {53}^o,\tan {37}^o,\tan {53}^o$ in terms of fraction.

Answer 1

Step 1: Construct the right-angled triangle using the Pythagorean triplets

Construct the right-angled triangle $△ABC$, with$\mathrm{\angle }ACB={37}^o,\mathrm{\angle }ABC={90}^o,\mathrm{\angle }BAC=53^o$ and sides as$3cm,4cm$and $5cm$.

Step 2: Find the value of given trigonometric function in terms of fraction

We know that, $\sin \theta ={\displaystyle \frac{P}{H}}$ , and for angle$\theta ={37}^o$ , $P=3$ and $H=5$ .

So, $\sin {37}^o=\frac{3}{5}$ .
Similarly, for angle $\theta ={53}^o$ , $P=4$ and $H=5$ .

So,$\sin {53}^o=\frac{4}{5}$ .
Now, for angle $\theta ={37}^o$ , $P=3$and $B=4$ .

So, $\tan {37}^o=\frac{3}{4}$.
Similarly, for angle $\theta ={53}^o$ , $P=4$ and $B=3$ .

So, $\tan {53}^o=\frac{4}{3}$

Hence, $\sin {37}^o=\frac{3}{5},\sin {53}^o=\frac{4}{5},\tan {37}^o=\frac{3}{4},$and $\tan {53}^o=\frac{4}{3}$.