Question

In the triangle above, the sine of $x^\circ$ is 0.6. What is the cosine of $y^\circ$ ?

Answer 1

Hint : We know, sine and cosine are the ratios of side to the hypotenuse of a right angled triangle. So, to solve this problem, we have to find the sine of $x^\circ$ that resembles the ratio of which side to the hypotenuse of the triangle. Then we will find the cosine of $y^\circ$ resembles the ratio of which side to the hypotenuse. Then by using the value for sine of $x^\circ$ as given, we can find the required value, that is cosine of $y^\circ$ .

Complete step by step solution:

So, let us name the sides of the triangle as $A,B,C$ , such that the right angle is at $B$ .
The angle $x^\circ$ is at $C$ and the angle $y^\circ$ is at $A$ .
Now, we know, the sine of an angle is the ratio of the perpendicular to the hypotenuse in a right angled triangle.
Therefore, sine of $x^\circ$ can be written as, $\sin x = \dfrac{{AB}}{{AC}} = 0.6$
Also, we know, cosine of an angle is the ratio of the base to the hypotenuse in a right angled triangle.
Therefore, we can write, cosine of $y^\circ$ as, $\cos y = \dfrac{{AB}}{{AC}}$
Therefore, it is clearly visible to us that, $\sin x = \cos y = \dfrac{{AB}}{{AC}}$ .
So, we get a cosine of $y^\circ$ as, $\cos y = \dfrac{{AB}}{{AC}} = 0.6$ .
So, the correct answer is “0.6”.

Note : We can also solve this problem in another way that is, if the sine of an angle is $0.6 = \dfrac{3}{5}$ , then the angle is $37^\circ$ , therefore, the other angle of the triangle other than right angle is clearly $53^\circ$ . Therefore, the cosine of $53^\circ$ is also $0.6 = \dfrac{3}{5}$ . The formulae of the trigonometric functions must be clearly understood before attempting such questions.