# Trigonometric Ratios

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## Introduction:

Consider $$\Delta ABC$$ which is right-angled at $$b$$. $$\angle A$$ is equal to $$\theta$$: Here, $$\angle A$$ is an acute angle. Note the position of the side $$BC$$ with respect to $$\angle A$$. It faces $$\angle A$$.
We call it the side opposite to angle A or Perpendicular ($$p$$).

$$AC$$ is the hypotenuse ($$h$$) of the right triangle $$ABC$$.

The side $$AB$$ is a part of $$\angle A$$.
So, we call it the side adjacent to angle A or Base ($$b$$).

✍Note: The position of perpendicular and base change when you consider angle $$C$$ in place of $$A$$, but the position of the hypotenuse will remain the same.

## What are Trigonometric Ratios?

In a right-angled triangle, how many different ratios of sides $$\left( {p,{\text{ }}b,{\text{ and }}h} \right)$$ can we make in total? ✍Note: In a ratio, we can take two sides at a time from total of three sides of a right-angled triangle $$\left( {p,{\text{ }}b,{\text{ and }}h} \right)$$.

The six possible ratios can be defined for $$\angle A$$ (or $$\theta$$):

•     \begin{align} \text {sine of}\; \rm{\theta} \;or\; \text {sin} \;\rm{\theta} = \frac{p}{h}\end{align}

•     \begin{align} \rm{cosine \,of\, } \theta \,or\,\rm{\cos}\,\theta = \frac{b}{h}\end{align}

•     \begin{align}\rm{tangent \,of} \,\theta \,\rm{or \,\tan }\,\theta = \frac{p}{h}\end{align}

•      \begin{align}\rm{cosecant \,of }\,\theta \,\rm{or\, cosec} \,\theta = \frac{h}{p}\end{align}

•      \begin{align}\rm{secant ,of}\,\theta \,\rm{or\,\sec}\,\theta = \frac{h}{b}\end{align}

•       \begin{align}\rm{cot angent \,of }\,\theta \,\rm{or\,\cot }\,\theta = \frac{b}{p}\end{align}

These six ratios are called as Trigonometric Ratios.

## Reciprocal relations for Trigonometric ratios:

•   \begin{align}\sin \theta = \frac{1}{{\rm{cosec}\;\theta }}\,\,;\,\sin \theta \,\rm{cosec}\;\theta = 1\end{align}

•   \begin{align}\cos \theta = \frac{1}{{\sec \theta }}\,\,;\,\cos \theta \,\sec \theta = 1\end{align}

•   \begin{align}\tan \theta = \frac{1}{{\cot \theta }}\,;\,\tan \theta \,\cot \theta = 1\end{align}

We also note the following:

• \begin{align}\tan \theta = \frac{p}{b} =\frac{\frac{p}{h}}{\frac{b}{h}} = \frac{{\sin \theta }}{{\cos \theta }}\end{align}
• \begin{align}\cot \theta = \frac{1}{{\tan \theta }} = \frac{{\cos \theta }}{{\sin \theta }}\end{align}

✍Note: We can write all of $$\tan \theta ,\;\cot \;\theta, \rm{cosec}\; \theta$$ and $$\sec \theta$$ in terms of $$\sin \theta$$ and $$\cos \theta$$.

## Solved Examples:

Example 1: In a $$\Delta ABC$$, $$\angle A = {60^\circ}$$and $$\angle B = {90^\circ }$$. Also, $$AC$$ = 4 cm. It is known that $$\sec {60^ \circ} = 2$$. Find the lengths of $$AB$$ and $$BC$$.

Solution: Observe the following figure: We have,

\begin{align} \sec \theta &= \frac{h}{b} \hfill \\ \Rightarrow \sec {60^\circ } &= \frac{4}{b} \hfill \\ \Rightarrow b &= \frac{4}{{\sec {{60}^\circ }}} = \frac{4}{2} \hfill \\ \end{align}

$\Rightarrow \boxed{AB = b = 2\;{\text{cm}}}$

Using the Pythagoras theorem, we have

\begin{align} p &= \sqrt {{h^2} - {b^2}} \hfill \\ &= \sqrt {16 - 4} \hfill \\ &= \sqrt {12} \hfill \\ \end{align}

$\Rightarrow \boxed{BC = p = 2\sqrt 3 {\text{ }}cm}$

Example 2: $$\Delta PQR$$ is right-angled at $$Q$$, and $$\angle R$$ equals $${60^\circ}$$. It is known that $$\tan {60^\circ } = \sqrt 3$$. If $$PR$$ = 10 cm, Find the lengths of $$PQ$$ and $$QR$$.

Solution: Observe the following figure: We have

\begin{align}&\tan {60^\circ } = \sqrt 3 = \frac{p}{b} \Rightarrow p = b\sqrt 3 \end{align}

Also, using the Pythagoras Theorem:

\begin{align} {p^2} + {b^2} &= {10^2} \hfill \\ \Rightarrow 3{b^2} + {b^2} &= 100 \hfill \\ \Rightarrow 4{b^2} &= 100 \hfill \\ \Rightarrow {b^2} &= 25 \hfill \\ \end{align}

$\Rightarrow \boxed{QR = b = 5cm}$

Finally,

$p = b\sqrt 3$

$\Rightarrow \boxed{PQ = p = 5\sqrt 3 {\text{ }}cm}$

Example 3: It is known that $$\sin {60^\circ } = \frac{{\sqrt 3 }}{2}$$ from this information, deduce the value of $$\cos {60^\circ }$$.

Solution: Consider the following figure: We have

\begin{align} \sin {60^0} &= \frac{p}{h} = \frac{{\sqrt 3 }}{2} \hfill \\ \Rightarrow p &= \frac{{\sqrt 3 h}}{2} \hfill \\ \end{align}

Now, using the Pythagoras theorem, we have,

\begin{align} {p^2} + {b^2} &= {h^2} \hfill \\ \Rightarrow \frac{{3{h^2}}}{4} + {b^2} &= {h^2} \hfill \\ \Rightarrow {b^2} &= {h^2} - \frac{{3{h^2}}}{4} \hfill \\ \Rightarrow {b^2} &= \frac{{{h^2}}}{4} \hfill \\ \Rightarrow \frac{b}{h} &= \frac{1}{2} \hfill \\ \end{align}

Thus,

$\boxed{\cos {{60}^\circ } = \frac{b}{h} = \frac{1}{2}}$ Challenge: In example 3, we evaluated the value of $$\cos {60^\circ }$$, similarily, can you evaluate the values of $$\sec \theta$$, $$\cot \theta$$, $$\tan \theta$$, and $${\text{cosec }}\theta$$.

⚡Tip: Use reciprocal relations of trigonometric ratios.

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