Question

If \[\cos A = \dfrac{5}{{13}}\], how do you find \[\sin A\] and \[\tan A\]?

Answer 1

Hint: In the given question, we have been given the value of a trigonometric function. We have to calculate the values of two other trigonometric functions. To do that, we are going to use the basic formula of the given trigonometric function in terms of hypotenuse, perpendicular, and base. Then we are going to use the values into the formulae of the other two functions.

Formula Used:
We are going to use the following formulae:
\[\cos \theta = \dfrac{B}{H},\sin \theta = \dfrac{P}{H},\tan \theta = \dfrac{P}{B}\]

Complete step by step answer:
We are given that, \[\cos A = \dfrac{5}{{13}}\].
Now, we know that \[\cos \theta = \dfrac{B}{H}\].
Hence, we can say that for some \[x\], \[B = 5x\] and \[H = 13x\].
So, applying Pythagoras theorem,
\[P = \sqrt {{H^2} - {B^2}} \]
So, \[P = \sqrt {{{\left( {13x} \right)}^2} - {{\left( {5x} \right)}^2}} = \sqrt {169{x^2} - 25{x^2}} = \sqrt {144{x^2}} = 12x\]
So, for \[\sin A\], we have,
\[\sin A = \dfrac{P}{H} = \dfrac{{12x}}{{13x}} = \dfrac{{12}}{{13}}\]
and, for \[\tan A\], we have,

\[\tan A = \dfrac{P}{B} = \dfrac{{12x}}{{5x}} = \dfrac{{12}}{5}\]

Note:
In this question we had been given the value of one basic trigonometric function. We had to calculate the value of two other basic trigonometric functions. We did that by first expressing the given trigonometric function in terms of hypotenuse, perpendicular, and base, then we found the relative value of one missing parameter. So, we had the three values for any basic trigonometric function. Then we used these parameters to find the value of the other two basic trigonometric functions. So, it is necessary that we know the formulae of all the basic trigonometric functions.