Question

How do you find the values of all six trigonometric functions of a right triangle $ABC$, where $C$ is the right angle, given $a = 1,$ $b = 3$?

Answer 1

In this question, we need to calculate the values of six trigonometric functions for a given right angled triangle. Firstly, we will draw a diagram of the right angled triangle for the given data. We will consider the two vertices other than the right angle vertex and we will write the adjacent, opposite sides to the respective vertex. Now we will make use of basic definitions of the trigonometric ratios and calculate the values for the six trigonometric functions.

Complete step-by-step solution:
Given that the triangle $ABC$ is a right angled triangle with $C$ as the right angle. Given the sides as
$a = 1,$ $b = 3$.
We will consider $a = 1$ as adjacent side and $b = 3$ as opposite of the triangle.
To find the hypotenuse say $c$, we use the Pythagorean theorem.
We know that in right angled triangle,
${(adjacent)^2} + {(opposite)^2} = {(hypotenuse)^2}$
i.e. ${a^2} + {b^2} = {c^2}$
$\Rightarrow {1^2} + {3^2} = {c^2}$
$\Rightarrow 1 + 9 = {c^2}$
$\Rightarrow 10 = {c^2}$
Now we isolate c by taking square root on both sides, we get,
$\Rightarrow c = \sqrt {10}$
Hence the triangle $ABC$ can be drawn as below.

In the above triangle, the hypotenuse $AB = c = \sqrt {10}$
Now considering the angle $\angle BAC$ in the above triangle. Adjacent side to the angle $\angle BAC$ is $AC = b = 3$ and opposite side to the angle $\angle BAC$ is $BC = a = 1$.
From the basic definitions of trigonometric ratios, we have the values of all trigonometric ratios as,
$\sin A = \dfrac{{{\text{opposite(a)}}}}{{{\text{hyotenuse(c)}}}} = \dfrac{1}{{\sqrt {10} }}$
$\cos A = \dfrac{{{\text{adjacent(b)}}}}{{{\text{hyotenuse(c)}}}} = \dfrac{3}{{\sqrt {10} }}$
$\tan A = \dfrac{{{\text{opposite(a)}}}}{{{\text{adjacent(b)}}}} = \dfrac{1}{3}$
$\cot A = \dfrac{{{\text{adjacent(b)}}}}{{{\text{opposite(a)}}}} = \dfrac{3}{1} = 3$
$\sec A = \dfrac{{{\text{hypotenuse(c)}}}}{{{\text{adjacent(b)}}}} = \dfrac{{\sqrt {10} }}{3}$
$\csc A = \dfrac{{{\text{hypotenuse(c)}}}}{{{\text{opposite(a)}}}} = \dfrac{{\sqrt {10} }}{1} = \sqrt {10}$
Now consider the angle $\angle ABC$ in the above triangle. Adjacent side to the angle $\angle ABC$ is $BC = a = 1$ and opposite side to the angle $\angle ABC$ is $AC = b = 3$.
From the basic definitions of trigonometric ratios, we have the values of all trigonometric ratios as,
$\sin B = \dfrac{{{\text{opposite(b)}}}}{{{\text{hyotenuse(c)}}}} = \dfrac{3}{{\sqrt {10} }}$
$\cos B = \dfrac{{{\text{adjacent(a)}}}}{{{\text{hyotenuse(c)}}}} = \dfrac{1}{{\sqrt {10} }}$
$\tan B = \dfrac{{{\text{opposite(b)}}}}{{{\text{adjacent(a)}}}} = \dfrac{3}{1} = 3$
$\cot B = \dfrac{{{\text{adjacent(a)}}}}{{{\text{opposite(b)}}}} = \dfrac{1}{3}$
$\sec B = \dfrac{{{\text{hypotenuse(c)}}}}{{{\text{adjacent(a)}}}} = \dfrac{{\sqrt {10} }}{1} = \sqrt {10}$
$\csc B = \dfrac{{{\text{hypotenuse(c)}}}}{{{\text{opposite(b)}}}} = \dfrac{{\sqrt {10} }}{3}$

Note: In this problem, we have a lot of data related to the triangle. So, it is easy to solve when we have the diagram representation of the given data. If we have a triangle and its sides, it will be easier to calculate the trigonometric ratios. Without representing the triangle, it will be difficult to solve the given problem. And we must remember the all the representation of trigonometric ratios and must be careful while substituting the values.