Question

For a right triangle ABC, how do you find the sine, cosine and tangent of angle A?

Answer 1

Hint: Assume a right-angle triangle ABC with right angle at B. Consider a, b and c as the sides opposite to the angles A, B and C respectively of the right triangle ABC. Use the relations: - \[\sin A=\dfrac{P'}{H'}\], \[\cos A=\dfrac{B'}{H'}\] and \[\tan A=\dfrac{P'}{B'}\], where P’ = perpendicular, B’ = base and H’ = hypotenuse to get the relation among the angles and sides. Consider the side opposite to angle A as the perpendicular.

Complete step by step answer:
Here, we have been asked to find the sine, cosine and tangent of angle A for a right-angle triangle.
Now, let us assume a right-angle triangle ABC has a right-angle at B. Let us draw a rough diagram for the triangle.

In the above figure of the right-angle triangle, according to the convention we have assumed a, b and c as the sides opposite to the angles A, B and C respectively. Now, we have to find three trigonometric ratios of angle A. So, if we will consider the angle A then according to the convention we have to assume the side ‘a’ as the perpendicular. Therefore, the side ‘i’ will be referred to as the base because side ‘b’ will always be the hypotenuse. So, considering P’ = perpendicular, B’ = base and H’ = hypotenuse, we have,
\[\Rightarrow BC=a=P',AB=c=B'\] and \[AC=b=H'\]
For a right-angle triangle ABC,
(i) \[\sin A=\dfrac{P'}{H'}\]
\[\begin{align}
& \Rightarrow \sin A=\dfrac{a}{b} \\
& \Rightarrow \sin A=\dfrac{BC}{AC} \\
\end{align}\]
(ii) \[\cos A=\dfrac{B'}{H'}\]
\[\begin{align}
& \Rightarrow \cos A=\dfrac{c}{b} \\
& \Rightarrow \cos A=\dfrac{AB}{AC} \\
\end{align}\]
(iii) \[\tan A=\dfrac{P'}{B'}\]
\[\begin{align}
& \Rightarrow \tan A=\dfrac{a}{c} \\
& \Rightarrow \tan A=\dfrac{BC}{AB} \\
\end{align}\]
Hence, the above three relations are our answers.

Note: One may note that there are three more trigonometric ratios, namely: - secant, cosecant and cotangent. We can also find them using the three relations: - \[\sec A=\dfrac{1}{\cos A}\], \[\csc A=\dfrac{1}{\sin A}\] and \[\cot A=\dfrac{1}{\tan A}\]. You must remember the naming conventions used for any triangle so that you may not get confused. Do not forget to draw a rough diagram of the right-angle triangle for these types of questions. Note that if you will consider angle A equal to 90 degrees for the above question then you will get the value of the tangent function equal to infinite because the base length will become 0.