CBSE Class 10 Chapter 8 Introduction of Trigonometry Examples

 

Example 1.

Given $\tan A =\frac {4}{3}$, find the other trigonometric ratios of the angle A.

Solution.

Let's take a $\triangle ABC$ Where B is a right angle and $\angle A$ is a acute angle. We know that  $\tan A = \frac {\text {Opposite side to angle A}}{\text {adjacent side to angle A }}=\frac {BC}{AB}=\frac {4}{3}$ Therefore, if $BC = 4k$ then $AB = 3k$. Here k is a positive number. By using Pythagoras Theorem, we can find AC (hypotenuse), $AC^{2} = AB^{2} + BC^{2}$ $AC^{2} = (3k)^{2} + (4k)^{2}$ $AC^{2} = 9k^{2} + 16k^{2}$ $AC^{2} = 25k^{2}$ $AC = 5k$ Now that we have all three sides of the right-angled triangle; AC = 5k, AB=3k, BC=4k, so $\sin A = \frac {BC}{AC}=\frac {4k}{5k}=\frac {4}{5}$ $\cos A = \frac {AB}{AC}=\frac {3k}{5k}=\frac {3}{5}$ and their reciprocals are $\text {cosec } A = \frac {1}{\sin A}= \frac {1}{\frac {4}{5}}=\frac {5}{4}$ $\sec A = \frac {1}{\cos A}= \frac {1}{\frac {3}{5}}=\frac {5}{3}$ $\cot A = \frac {1}{\tan A}= \frac {1}{\frac {4}{3}}=\frac {3}{4}$

Example 2.

If $\angle B$ and $\angle Q$ are acute angles such that $\sin B = \sin Q$, then prove that $\angle B = \angle Q$.

Solution.

Let's take two triangle ABC and PQR such that $\sin B = sin Q$ As per construction, we have $\sin B = \frac {AC}{AB}$ and $\sin Q = \frac {PR}{PQ}$ As per given, $$\frac {AC}{AB}=\frac {PR}{PQ}$$ Therefore, we can say $$\frac {AC}{PR}=\frac {AB}{PQ}=k .........................(i)$$ In $\triangle ABC$ by using Pythagoras theorem: $AB^{2} = AC^{2} + BC^{2}$ $BC = \sqrt {AB^{2}-AC^{2}}$ In $\triangle PQR$ by using Pythagoras theorem: $PQ^{2} = PR^{2} + QR^{2}$ $QR = \sqrt {PQ^{2}-PR^{2}}$ So, \(\frac {BC}{QR} = \frac {\sqrt {AB^{2}-AC^{2}}}{\sqrt {PQ^{2}-PR^{2}}} from equation (i) $\frac {AC}{PR}=k \text { or } AC = k.PR$ $\frac {AB}{PQ}=k \text { or } AB = k.PQ$ by putting these value So, $\frac {BC}{QR} = \frac {\sqrt {k^{2}.PQ^{2}-k^{2}.PR^{2}}}{\sqrt {PQ^{2}-PR^{2}}}=\frac {k\sqrt {PQ^{2}-PR^{2}}}{\sqrt {PQ^{2}-PR^{2}}}=k....................... (ii)$ By equation (i) and (ii) we have $\frac {AC}{PR}=\frac {AB}{PQ}=\frac {BC}{QR}$ We know that two triangles are similar, if their corresponding sides are in the same ratio (or proportion), so we can conclude that triangles ABC and PQR are similar. $\triangle ABC \sim \triangle PQR$ and therefore $\angle B = \angle Q$.

Example 3.

Consider $\triangle ACB$, right-angled at $C$, in which $AB = 29$ units, $BC = 21$ units and $angle ABC = \theta$ (see Fig. 8.10). Determine the values of  (i) $\cos^{2}\theta +\sin^{2}\theta$ (ii) $\cos^{2}\theta -\sin^{2}\theta$

Solution.

As per shown in figure, In $\triangle ABC$ , we have $AC = \sqrt{AB^{2}-BC^{2}}=\sqrt{(29)^{2}-(21)^{2}}$ $AC = \sqrt{(29-21)(29+21)}=\sqrt{(8)(50)}=\sqrt{400}=\text {20 unit}$ So, By using trigonometric ratio $\sin \theta = \frac {AC}{AB}=\frac {20}{29}$ $\cos \theta = \frac {BC}{AB}=\frac {21}{29}$ Hence, (i) $\cos^{2}\theta +\sin^{2}\theta=\left(\frac {20}{29}\right)^{2}+\left(\frac {21}{29}\right)^{2}=\frac {20^{2}+21^{2}}{29^{2}}=\frac {400+441}{841}=1$ (ii) $\cos^{2}\theta -\sin^{2}\theta=\left(\frac {21}{29}\right)^{2}-\left(\frac {20}{29}\right)^{2}=\frac {(21+20)(21-20)}{29^{2}}=\frac {41}{841}$

Example 4.

In a right triangle $ABC$, right-angled at $B$, if tan $A = 1$, then verify that $2\sin A \cos A = 1$.

Solution.

As per given, In $\triangle ABC$, $\tan = \frac {AB}{BC}=1$ So, we can say that the length of the side opposite $\angle A$ is equal to the length of the adjacent side. Therefore, $AB = BC$ Let $AB = BC = k$, where $k$ is a positive number. Now, in a right-angled triangle, we can use the Pythagorean theorem: $AC = \sqrt {AB^{2}+BC^{2}}$ Putting value $AC = \sqrt {k^{2}+k^{2}}=k\sqrt{2}$ Therefore, $\sin A = \frac {BC}{AC}=\frac {1}{\sqrt {2}}$ and $\cos A = \frac {AB}{AC}=\frac {1}{\sqrt {2}}$ Hence, $2\sin A \cos A = 2\left(\frac {1}{\sqrt {2}}\right)\left(\frac {1}{\sqrt {2}}\right)=1$

Example 5.

In \triangle OPQ\), right-angled at $P$,$OP = 7$ cm and $OQ – PQ = 1$ cm (see Fig. 8.12). Determine the values of $\sin Q$ and $cos Q$.

Solution.

As per given, In $\triangle OPQ$, $OP = 7$ cm $OQ-PQ = 1$ or $OQ = 1+PQ$ Now, in a right-angled triangle, we can use the Pythagorean theorem: $OQ^{2}=OP^{2}+PQ^{2}$ As per given, putting $OQ$ value $(1+PQ)^{2}=OP^{2}+PQ^{2}$ We know that $(a+b)^{2}=a^{2}+b^{2}+2ab$ so, $1^{2}+PQ^{2}+2PQ=OP^{2}+PQ^{2}$ Divide both side by $PQ^{2}$ and $OP$ value as per given, we get $1+2PQ = 7^{2}$ $2PQ = 49-1$ $2PQ = 48=24$ cm. So, $PQ = 24 \text { cm}$ and $OQ = 1+PQ = 25 \text { cm}$ Therefore,  $\sin Q = \frac {OP}{OQ}=\frac {7}{25}$ $\cos Q = \frac {PQ}{OQ}=\frac {24}{25}$

Example 6.

In $\triangle ABC$, right-angled at $B$, $AB = 5\text{ cm}$ and $\angle ACB = 30^{0}$ (see Fig. 8.19). Determine the lengths of the sides $BC$ and $AC$.

Solution.

$\text{In } \triangle ABC$, right-angled at $B$, $AB = 5\, \text{cm}$ and $\angle ACB = 30^\circ$. We have to find the lengths of the sides $BC$ and $AC$. 1. By using Sine Ratio: $\sin 30^\circ = \frac{AB}{AC}$ by putting $\sin 30^\circ = \frac {1}{2}$ $\frac{1}{2} = \frac{5}{AC}$ $AC = \frac{5}{1/2} = 10\, \text{cm}$ 2. By using Cosine Ratio: $\cos 30^\circ = \frac{BC}{AC}$ by putting $\cos 30^\circ = \frac {\sqrt {3}}{2}$ $\frac{\sqrt{3}}{2} = \frac{BC}{10}$ $BC = \frac{\sqrt{3}}{2} \times 10 = 5\sqrt{3}\, \text{cm}$ Therefore, the lengths of the sides $BC$ and $AC$ are $5\sqrt{3}\, \text{cm}$ and $10\, \text{cm}$, respectively.

Example 7.

In $\triangle PQR$, right-angled at $Q$ (see Fig. 8.20), $PQ = 3 \text { cm}$ and $PR = 6\text { cm}$. Determine $\angle QPR$ and $\angle PRQ$.

Solution.

Given: $\text{In } \triangle PQR$, right-angled at $Q$, $PQ = 3 \, \text{cm}$ and $PR = 6 \, \text{cm}$. We have to find  $\angle QPR$ and $\angle PRQ$. 1. By using Sine Ratio: We know that  $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}$ putting value $\sin \angle PRQ = \frac{PQ}{PR} = \frac{3}{6} = \frac{1}{2}$ To find $\angle PRQ$, take the inverse sine (arcsin) of $\frac{1}{2}$: $\angle PRQ = \arcsin\left(\frac{1}{2}\right)$ In degrees, this is equal to $30^\circ$. 2. We know that sum of angle in a right angle triangle is $180^\circ$. so $\angle PQR + \angle QPR + \angle PRQ = 180^\circ$ Putting value $90^\circ + \angle QPR + 30^\circ = 180^\circ$ $\angle QPR = 60^\circ$ Therefore, in $\triangle PQR$, right-angled at $Q$: $\angle QPR = 60^\circ$ $\angle PRQ = 30^\circ$

Example 8.

If $\sin (A-B)=\frac{1}{2}$, $\cos (A+B)=\frac {1}{2}$, $0^\circ < A+B \leq 90^\circ$, $A > B$, Find $A$ and $B$.

Solution.

Given: $\sin (A-B)=\frac{1}{2}$, $\cos (A+B)=\frac {1}{2}$ and $0^\circ < A+B \leq 90^\circ$ We know that- $\sin 30^\circ = \frac {1}{2}$ So, we can write as $\sin (A-B)=\sin 30^\circ$ $A-B= 30^\circ ...................... (i)$ We also know that- $\cos 60^\circ = \frac {1}{2}$ So, we can write as $\cos (A+B)=\cos 60^\circ$ $A+B= 60^\circ ...................... (i)$ By adding equation $(i)$ and $(ii)$ $A-B+A+B = 30 + 60$ $2A = 90$ $A = 45^\circ$ By putting $A=45^\circ$ in equation (i) $45^\circ-B= 30^\circ$ $45^\circ-30^\circ= B$ $B = 15^\circ$ So the solution is $A=45^\circ$ and $B=15^\circ$ which satisfying given condition.

Example 9.

Express the ratios $\cos A$, $\tan A$ and $\sec A$ in terms of $\sin A$.

Solution.

We know that, Pythagorean identity is: $\cos^{2}A + \sin^{2}A = 1$ or $\cos^{2}A = 1-\sin^{2}A$ or $\cos A = \pm \sqrt {1-\sin^{2}A}$ Here A is acute angle when \(A<90|) so, it is a positive angle, then $\cos A =  \sqrt {1-\sin^{2}A}...............(i)$ Ratio of $\bf \sec A$ $\sec A = \frac {1}{\cos A}$ Putting $\cos A$ value from equation $(i)$ $\sec A = \frac {1}{\sqrt {1-\sin^{2}A}}$ Ratio of $\bf \tan A$ $\tan A = \frac {\sin A}{\cos A}$ Putting $\cos A$ value from equation $(i)$ $\tan A = \frac {\sin A}{{\sqrt {1-\sin^{2}A}}}$

Ex. 10.

Prove that $\sec A (1-\sin A)(\sec A + tan A) = 1$.

Solution.

We know that $\sec A = \frac {1}{\cos A}$ and $\tan A =\frac {sin A}{cos A}$ $\sec A (1-\sin A)(\sec A + \tan A)$  Putting $\sec A$ and $\tan A value$ $= \frac{1}{\cos A} (1 - \sin A)\left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$ $=\frac{(1 - \sin A)}{\cos A} \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right)$ $=\frac{(1 - \sin A)}{\cos A} \left(\frac{1+ sin A}{\cos A} \right)$ $=\frac{(1 - \sin A)(1+ sin A)}{\cos A\times \cos A}$ $=\frac{(1 - \sin A)(1 + \sin A)}{\cos^2 A}$ Since, we know that $a^{2}-b^{2}=(a-b)(a+b)$ $=\frac{1 - \sin^2 A}{\cos^2 A}$ We also know that $\cos^{2} A+\sin^{2} A = 1$ or $\cos^{2} A= 1-\sin^{2} A$ So we can write as $=\frac{\cos^2 A}{\cos^2 A}$ $=1$ Therefore, $\sec A (1-\sin A)(\sec A + \tan A) = 1$. Its Proved.

Ex. 11.

Prove that $\frac {\cot A - \cos A}{\cot A + \cos A} = \frac {\text {cosec }A-1}{\text {cosec }A+1}$.

Solution.

$$\begin{align*} \text{LHS} &= \frac{\cot A - \cos A}{\cot A + \cos A} \\ &= \frac{\frac{\cos A}{\sin A} - \cos A}{\frac{\cos A}{\sin A} + \cos A} \\ &= \frac{\frac{\cos A - \sin A \cdot \cos A}{\sin A}}{\frac{\cos A + \sin A \cdot \cos A}{\sin A}} \\ &= \frac{\cos A(1 - \sin A)}{\cos A(1 + \sin A)} \\ &= \frac{1 - \sin A}{1 + \sin A}\\ &= \frac {\frac {(1-\sin A)}{\sin A}}{\frac {(1+\sin A)}{\sin A}} \\ =& \frac {\frac {1}{\sin A}-\frac {\sin A}{\sin A}}{\frac {1}{\sin A}+\frac {\sin A}{Sin A}} \\ &= \frac {\frac {1}{sin A}-1}{\frac {1}{sin A}+1} \\ &= \frac {\text {cosec }A-1} {\text {cosec }A+1}. \end{align*}$$ Therefore, $\frac{\cot A - \cos A}{\cot A + \cos A} = \frac{\text {cosec }A - 1}{\text {cosec }A + 1}$.

Ex. 12.

Prove that $\frac {\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac {1}{\sec\theta-\tan\theta}$, using the identity $\sec^{0}\theta = 1+tan^{0}\theta$.

Solution.

$$\begin{align*} \text{LHS} &=\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} \\ &= \left(\frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}\right)\times \left(\frac{\sin\theta+\cos\theta+1}{\sin\theta+\cos\theta+1}\right) \\ &= \left(\frac{\sin\theta+1-\cos\theta}{\sin\theta+\cos\theta-1}\right)\times \left(\frac{\sin\theta+1+\cos\theta}{\sin\theta+\cos\theta+1}\right) \\ &= \frac {(\sin\theta+1)^{2}-\cos^{2}\theta}{(\sin\theta+\cos\theta)^{2}-1^{2}}\\ &=\frac {\sin^{2}\theta+1+2\sin\theta-\cos^{2}\theta}{\sin^{2}\theta+\cos^{2}\theta+2\sin\theta\cos\theta-1}\\ &= \text {We know that } \sin^{2}\theta+\cos^2\theta = 1 \text { and } \sin^{2}\theta=1-\cos^2\theta \\ &= \text {Therefore,}\\ &= \frac {1-\cos^{2}\theta+1+2\sin\theta-\cos^{2}\theta}{1+2\sin\theta\cos\theta-1}\\ &= \frac {2-2\cos^{2}\theta+2\sin\theta}{2\sin\theta\cos\theta}\\ &= \frac {\sin^{2}\theta+\sin\theta}{\sin\theta\cos\theta}\\ &= \frac {\sin\theta+1}{\cos\theta}\\ &= \frac {1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\\ &= \sec\theta + \tan\theta \\ &= \left(\sec\theta + \tan\theta\right) \times \left(\frac{\sec\theta-\tan\theta}{\sec\theta-\tan\theta}\right)\\ &=\frac {\sec^{2}\theta-\tan^{2}\theta}{\sec\theta-\tan\theta}\\ &=\text {As we know and given that} \sec^{2}\theta-\tan^{2}\theta=1 \\ &= \text {Therefore,}\\ &=\frac {1}{\sec\theta-\tan\theta}=RHS\\ &= \text {Hence, It's proved.} \end{align*}$$