Question

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

Answer 1

Hint: In this question, we have to write the given trigonometric function in terms of fraction.
We know, that in a right- angled triangle, there are three sides, perpendicular, base and the hypotenuse.
And, $\sin \theta ={\displaystyle \frac{P}{H}}$ , where, $P$ is the length of perpendicular and $H$ is referred to as the length of hypotenuse, whereas, $\tan \theta ={\displaystyle \frac{P}{B}}$ , where, $P$ is the length of perpendicular and $B$ is the length of base of the triangle.

Complete answer:
Given trigonometric functions $\sin {37}^o,\sin {53}^o,\tan {37}^o,\tan {53}^o$ .
To write these trigonometric functions in terms of fraction.
Consider a right- angled triangle, $△ABC$ , with $\mathrm{\angle }ACB={37}^o$ and $\mathrm{\angle }ABC={90}^o$ .

Then, we have, using angle sum property of a triangle, that, $\mathrm{\angle }ABC+\mathrm{\angle }BAC+\mathrm{\angle }ACB={180}^o$ , i.e., ${37}^o+{90}^o+\mathrm{\angle }BAC={180}^o$ . On solving, we get, $\mathrm{\angle }BAC={180}^o-{127}^o$ i.e., $\mathrm{\angle }BAC={53}^o$ .
Now, let length of side $AB$ is $3units$ and length of side $BC$ is $4units$ , then, by Pythagoras theorem, we have, $A{B}^2+B{C}^2=A{C}^2$ , putting values, we get, $3^2+4^2=9+16=25$ , hence, $AC=5units$ .
Now, we know, $\sin \theta ={\displaystyle \frac{P}{H}}$ , and for angle $\theta ={37}^o$ , $P=3$ and $H=5$ . So, $\sin {37}^o={\displaystyle \frac{3}{5}}$ .
Similarly, for angle $\theta ={53}^o$ , $P=4$ and $H=5$ . So, $\sin {53}^o={\displaystyle \frac{4}{5}}$ .
Now, for angle $\theta ={37}^o$ , $P=3$ and $B=4$ . So, $\tan {37}^o={\displaystyle \frac{3}{4}}$ .
Similarly, for angle $\theta ={53}^o$ , $P=4$ and $B=3$ . So, $$\tan {53}^o={\displaystyle \frac{4}{3}}$$

Note:
It is not necessary to choose the lengths of sides of the triangle to be $3units$ or $4units$ . We can choose the length of sides of the right- angled triangle by our choices.
If $x^2=a^2$ , then, taking square root on both sides, we get, $x=\pm a$ , but in this question, we are talking about length of sides and length can never be negative. Hence, we have taken only the positive one.
For any angle, say $\theta$ , the sides opposite to this angle will be the perpendicular side, whereas, the third side except for the hypotenuse, will be the base of the triangle.