Question

Given $\cos \alpha = \dfrac{1}{3}$, how do you find $\sin \alpha$?

Answer 1

We will first observe that the ratio of base and hypotenuse gives us the cosine of any angle and then use the given ratio and then use the Pythagorean theorem to find the perpendicular and find sine.

Complete step-by-step answer:
We are given that $\cos \alpha = \dfrac{1}{3}$ and we need to find $\sin \alpha$.
We know that cosine of any angle is given by the ratio of base and the hypotenuse of any right – angled triangle that is we have: $\cos x = \dfrac{B}{H}$.
Now, since we are given that $\cos \alpha = \dfrac{1}{3}$.
Therefore, we can assume that the base of this triangle is x units and hypotenuse is 3x units.
Now, we know that the Pythagorean theorem is given by: ${H^2} = {P^2} + {B^2}$, where H is the hypotenuse, B is the base and P is the perpendicular of the given right – angled triangle.
Using, this we have: ${(3x)^2} = {x^2} + {P^2}$
On simplifying the left hand side, we will then obtain the following expression:-
$\Rightarrow 9{x^2} = {x^2} + {P^2}$
Simplifying it further, we will then obtain the following expression:-
$\Rightarrow P = 2\sqrt 2 x$ units
Now, we also know that the ratio of perpendicular and the hypotenuse gives us the sine of any angle in a right – angled triangle that is we have: $\sin x = \dfrac{P}{H}$
Putting the values, we will then obtain: $\sin \alpha = \dfrac{{2\sqrt 2 x}}{{3x}}$
Crossing – off x from numerator and the denominator on the right hand side, we will then obtain the following equation:-

$\Rightarrow \sin \alpha = \dfrac{{2\sqrt 2 }}{3}$

Note:
The students must note that we have used x with 1 and 3 in the base and hypotenuse because we are just given the ratio, the base and hypotenuse may be some multiple of 1 and 3 as well instead of directly being 1 unit and 3 units respectively.
The students must commit to memory the following formulas:-
$\sin x = \dfrac{P}{H}$
$\cos x = \dfrac{B}{H}$
Pythagorean Theorem: In a right angled triangle, we have ${H^2} = {P^2} + {B^2}$, where H is the hypotenuse, B is the base and P is the perpendicular of the given right – angled triangle.