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 \(\theta \)):
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\(\begin{align} \text {sine of}\; \rm{\theta} \;or\; \text {sin} \;\rm{\theta} = \frac{p}{h}\end{align}\)
-
\(\begin{align} \rm{cosine \,of\, } \theta \,or\,\rm{\cos}\,\theta = \frac{b}{h}\end{align}\)
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\(\begin{align}\rm{tangent \,of} \,\theta \,\rm{or \,\tan }\,\theta = \frac{p}{h}\end{align}\)
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\(\begin{align}\rm{cosecant \,of }\,\theta \,\rm{or\, cosec} \,\theta = \frac{h}{p}\end{align}\)
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\(\begin{align}\rm{secant ,of}\,\theta \,\rm{or\,\sec}\,\theta = \frac{h}{b}\end{align}\)
-
\(\begin{align}\rm{cot angent \,of }\,\theta \,\rm{or\,\cot }\,\theta = \frac{b}{p}\end{align}\)
These six ratios are called as Trigonometric Ratios.
Reciprocal relations for Trigonometric ratios:
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\(\begin{align}\sin \theta = \frac{1}{{\rm{cosec}\;\theta }}\,\,;\,\sin \theta \,\rm{cosec}\;\theta = 1\end{align}\)
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\(\begin{align}\cos \theta = \frac{1}{{\sec \theta }}\,\,;\,\cos \theta \,\sec \theta = 1\end{align}\)
-
\(\begin{align}\tan \theta = \frac{1}{{\cot \theta }}\,;\,\tan \theta \,\cot \theta = 1\end{align}\)
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.