CBSE Class 10 Chapter 8 Introduction of Trigonometry Exercise 8.1

Q1.

In $\triangle ABC$, right-angled at $B$, $AB=24 \text { cm}$, $BC=7\text{ cm}$. Determine: $\text{(i)} \sin A, \cos A$ $\text{(ii)} \sin C, \cos C$

Ans.

Given: In $\triangle ABC$  $\angle B = 90^{0}$ $AB = 24 \text { cm}$ $BC = 7 \text { cm}$ $AC$ be the hypotenuse By using Pythagoras Theorem: $AC^2 = AB^2 + BC^2$ $AC^2 = 24^2 + 7^2$ $AC^2 = 576 + 49$ $AC^2 = 625 \, \text{cm}^2$ $AC = \sqrt{625} = 25$ Therefore, $AC = 25 \, \text{cm}$ (i) $\sin(A)$, $\cos(A)$: $\sin(A) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$ $\cos(A) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$ (ii) $\sin(C)$, $\cos(C)$: $\sin(C) = \frac{\text{Opposite side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$ $\cos(C) = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$

Q2.

In Fig. 8.13, find $\tan P – \cot R$.

Ans.

Given: In given $\triangle PQR$  $\angle Q = 90^{0}$ $PQ = 12 \text { cm}$ $PR = 13 \text { cm}$ By using Pythagoras Theorem: $PR^2 = QR^2 + PQ^2$ $13^2 = QR^2 + 12^2$ $169 = QR^2 + 144$ $169-144 = QR^2$ $QR^2 = 25$ $QR = \sqrt{25} = 5$ Therefore, $QR = 5 \, \text{cm}$ $\tan P = \frac {\text {opposite side to}\angle P}{\text {Adjacent side to}\angle P}=\frac{QR}{PQ}\frac {5}{12} .......... (i)$ $\tan R = \frac {\text {opposite side to}\angle R}{\text {Adjacent side to}\angle R}=\frac{PQ}{QR}\frac {12}{5} .......... (ii)$ We know that $\cot R$ is reciprocal of $\tan R$, so $\cot R = \frac {1}{\tan R}$ Putting value of $\tan R$ from equation (ii) $\cot R = \frac {1}{\frac {12}{5}}=\frac {5}{12}....... (iii)$ Thus,  $\tan P – \cot R$ Putting value from equation (i) and (iii) $=\frac{5}{12}-\frac {5}{12}=0$

Q3.

$\text {If} \sin A = \frac {3}{4} \text {, calculate }\cos A \text { and } \tan A$.

Ans.

Given: In given $\triangle ABC$  $\angle B = 90^{0}$ $\sin A = \frac {\text {Opposite side to } \angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {3}{4}$ Let $BC = 3k$ $AC = 4k$ By using Pythagoras Theorem: $AC^2 = AB^2 + BC^2$ $4k^2 = AB^2 + 3k^2$ $16k^{2} = AB^2 + 9k^{2}$ $16k^{2}-9k^{2} = AB^2$ $AB^2 =7k^2$ $AB = \sqrt{7k^2} = \sqrt{7}k$ Therefore, $AB =\sqrt{7}$ $\cos A = \frac {\text {Adjacent side of } \angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {\sqrt{7}k}{4k}=\frac {\sqrt{7}}{4}$ $\tan A = \frac {\text {Opposite side of } \angle A}{\text {Adjacent side of} \angle A}=\frac {BC}{AB}=\frac {3k}{\sqrt {7}k}=\frac {3}{\sqrt {7}}$ Alternative Method: If $\sin A = \frac {3}{4}$, we can use the Pythagorean identity to find $\cos A$ and then use the definition of $\tan A$. The Pythagorean identity is given by: $\sin^{2} A + \cos^{2} A = 1$ $\cos^{2} A = 1 - \sin^{2} A$ By putting $\sin A$ value: $\cos^{2} A = 1 - \left(\frac {3}{4}\right)^{2}$ $\cos^{2} A = 1 - \left(\frac {9}{16}\right)$ $\cos^{2} A = \frac {7}{16}$ $\cos A = \pm \frac {\sqrt{7}}{4}$ Since, $\sin A$ is positive and $\cos A$ is in the same quadrant as $\sin A$, we take the positive root. $\cos A = \frac {\sqrt{7}}{4}$ As we know that $\tan A = \frac {\sin A}{\cos A}$ by putting value $\tan A = \frac {\frac {3}{4}}{\frac {\sqrt {4}}{4}}$ $\tan A = \frac {3}{4}\times \frac {4}{\sqrt {7}}$ $\tan A = \frac {3}{\sqrt {7}}$

Q4.

Given $15 \cot A = 8$, find $\sin A$ and $sec A$.

Ans.

Given: $15 \cot A = 8$ $\cot A = \frac {8}{15}$ We know that  $\tan A = \frac {1}{\cot A}$ By putting $\cos A$ value $\tan A = \frac {1}{\frac {8}{15}}=\frac {15}{8}$ So, $\tan A = \frac {\text {Opposite side of }\angle A}{\text {Adjacent side of }\angle A}=\frac {15}{8}$ For find AC (Hypotenuse) we use Pythagoras Theorem: $AC^2 = AB^2 + BC^2$ $AC^2 = 8k^2 + 15k^2$ $AC^{2} = 64k^2 + 225k^{2}$ $AC^{2}=289k^{2}$ $AC = 17k$ Therefore, $AC =17k$ $\sin A = \frac {\text {Opposite side to }\angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {15k}{17k}=\frac {15}{17}$ $\cos A = \frac {\text {Adjacent side to }\angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {8k}{17k}=\frac {8}{17}$ We know that- $\sec A = \frac {1}{\cos A}= \frac {1}{\frac {8}{17}}=\frac {17}{8}$

Q5.

Given $\sec\theta = \frac {13}{12}$, calculate all other trigonometric ratios.

Ans.

Given: $\sec\theta = \frac {13}{12}$ We know that  $\cos\theta = \frac {1}{\sec\theta}$ By putting $\sec \theta$ value $\cos\theta = \frac {1}{\frac {13}{12}}=\frac {12}{13}$ We also know that Pythagorean identity is: $\sin^{2} \theta + \cos^{2} \theta = 1$ By putting $\cos\theta$ value $\sin^{2} \theta + \left(\frac {12}{13}\right)^{2} = 1$ $\sin^{2} \theta =1- \frac {144}{169}$ $\sin^{2} \theta = \frac {25}{169}$ $\sin \theta = \pm \frac {5}{13}$ Since, $\sec\theta >0$ in the given expression and $\sin\theta$ is positive. so, $\sin \theta = \frac {5}{13}$ The other trigonometric ratios are: $\tan\theta =\frac {\sin\theta}{\cos\theta}=\frac {\frac {5}{13}}{\frac {12}{13}}=\frac {5}{12}$ $\cot\theta = \frac {1}{\tan\theta}=\frac {1}{\frac{5}{12}}=\frac {12}{5}$ $\text {cosec}\theta = \frac {1}{\sin\theta}=\frac {1}{\frac{5}{13}}=\frac {13}{5}$ $\sec\theta = \frac {13}{12}\text { given}$

Q6.

If $\angle A$ and $\angle B$ are acute angles such that $\cos A = cos B$, then show that $\angle A = \angle B$.

Ans.

Given : $\angle A$ and $\angle B$ are acute angles. $\cos A = cos B$ We have to show that $\angle A = \angle B$. In right angle $\triangle ABC$ $\cos A = \frac {Adjacent side to \angle A}{Hypotenuse}=\frac {AC}{AB}$ $\cos B = \frac {Adjacent side to \angle B}{Hypotenuse}=\frac {BC}{AB}$ As per given that $\cos A = \cos B$ Therefore,  $\frac {AC}{AB}=\frac {BC}{AB}$ $\text {Cross multiply}$ $AC \times AB = BC \times AB$ $\text {divide both side by AB, we get}$ $AC = BC$ Because both ratios are equal, then the lengths of AC and BC are the equal and $\triangle ABC$ is a Isosceles triangle, then angles opposite those equal sides would be equal. So, $\angle A = \angle B$ Alternative solution: Let's take a triangle ABC in which CP ┴ AB As per given $\cos A = \cos B$ $\frac {AP}{AC}=\frac {BP}{BC}$ $\frac {AP}{BP}=\frac {AC}{BC}$ Let $\frac {AP}{BP}=\frac {AC}{BC}=x$ Then, $AP = x.BP ............ (i)$ $AC = x.BC ............ (ii)$ By using Pythagoras theorem, we get From $\triangle CAP$ $AC^{2} = AP^{2}+CP^{2}$ $CP^{2}=AC^{2}-AP^{2}........ (iii)$ From $\triangle CBP$ $BC^{2}=BP^{2}+CP^{2}$ $CP^{2}=BC^{2}-BP^{2}......... (iv)$ from equatio (iii) and (iv), we get $AC^{2}-AP^{2}=BC^{2}-BP^{2}$ Putting value from (i) and (ii) $(x.BC)^{2}-(x.BP)^{2}=BC^{2}-BP^{2}$ $x^{2}BC^{2}-x^{2}BP^{2}=BC^{2}-BP^{2}$ $x^{2}(BC^{2}-BP^{2})=BC^{2}-BP^{2}$ $x^{2}=\frac {BC^{2}-BP^{2}}{BC^{2}-BP^{2}}=1$ $x=1$ Putt this value in equation (ii) $AC = 1.BC=BC$ Because angles opposite those equal sides would be equal. So, $\angle A = \angle B$

Q7.

$\text {If } \cot\theta =\frac {7}{8}, \text {evalute;}$ $\text {(i)} \frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ $\text {(ii)} \cot^{2}\theta$

Ans.

$\text {Let's take a } \triangle ABC \text{, where } \angle B \text {is a right angle as per above figure.}$ $\text {We know that-}$ $\cot\theta = \frac {\text {Adjacent side to} \theta}{\text {Opposite side to}\theta}=\frac {AB}{BC}=\frac {7}{8}$ $\text {Let}$$AB = 7x$ $\text {and}$ $BC = 8x$, $\text {Here }$$x$ $\text {is a positive integer.}$ By Pythagoras theorem in $\triangle ABC$ $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=(7x)^{2}+(8x)^{2}$ $AC^{2}=49x^{2}+64x^{2}$ $AC^{2}=113x^{2}$ $AC=\sqrt {113x^{2}}$ $AC=\sqrt {113x}$ Therefore, $\sin\theta =\frac {Opposite side to \theta}{Hypotenuse}=\frac {BC}{AC}=\frac {8x}{\sqrt{113x}}=\frac {8}{\sqrt{113}}$ $\cos\theta =\frac {Adjacent side to \theta}{Hypotenuse}=\frac {AB}{AC}=\frac {7x}{\sqrt{113x}}=\frac {7}{\sqrt{113}}$ $\text {(i)} \frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ $\text {We know that-} (a+b)(a-b)=a^{2}-b^{2}$ so, $\frac {(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\frac{1-\sin^{2}\theta}{1-\cos^{2}\theta}$ $=\frac {1-\left(\frac{8}{\sqrt{113}}\right)^{2}}{1-\left(\frac {7}{\sqrt{113}}\right)^{2}}$ $=\frac {1-\frac{64}{113}}{1-\frac {49}{113}}$ $=\frac {\frac {49}{113}}{\frac {64}{113}}$ $=\frac {49}{64}$ $\text {(ii)} \cot^{2}\theta$ $\cot^{2}\theta = \left(\frac {7}{8}\right)^{2}=\frac {49}{64}$

Q8.

If $3\cot A = 4$, check whether $\frac {1-\tan^{2}A}{1+\tan^{2}A}=\cos^{2}A-\sin^{2}A$ or not.

Ans.

Given that: $3\cot A = 4$  or $\cot A = \frac {4}{3}$ We know that- $\tan A = \frac {1}{\cot A}=\frac {1}{\frac {4}{3}}=\frac {3}{4}$ $\text {Let's take a } \triangle ABC \text{, where } \angle B \text {is a right angle as per above figure.}$ $\text {We know that-}$ $\cot A = \frac {\text {Adjacent side to }\angle A}{\text {Opposite side to}\angle A}=\frac {AB}{BC}=\frac {4}{3}$ $\text {Let}$$AB = 4x$ $\text {and}$ $BC = 3x$, $\text {Here }$$x$ $\text {is a positive integer.}$ By Pythagoras theorem in $\triangle ABC$ $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=(4x)^{2}+(3x)^{2}$ $AC^{2}=16x^{2}+9x^{2}$ $AC^{2}=25x^{2}$ $AC=\sqrt {25x^{2}}$ $AC=\sqrt {5x}$ Therefore, $\sin A =\frac {\text {Opposite side to} \angle A}{Hypotenuse}=\frac {BC}{AC}=\frac {3x}{5x}=\frac {3}{5}$ $\cos A =\frac {\text {Adjacent side to} \angle A}{Hypotenuse}=\frac {AB}{AC}=\frac {4x}{5x}=\frac {4}{5}$ So, $L.H.S. = \frac {1-\tan^{2}A}{1+\tan^{2}A}$ $=\frac{1-\left(\frac {3}{4}\right)^{2}}{1+\left(\frac {3}{4}\right)^{2}}=\frac {1-\frac {9}{16}}{1+\frac {9}{16}}=\frac {16-9}{16+9}=\frac {7}{25}$ $R.H.S. = \cos^{2}A-\sin^{2}A$ $\left(\frac {4}{5}\right)^{2}-\left(\frac {3}{5}\right)^{2}=\frac {16}{25}-\frac {9}{25}=\frac {16-9}{25}=\frac {7}{25}$ Hence, $\frac {1-\tan^{2}A}{1+\tan^{2}A}=\cos^{2}A-\sin^{2}A$

Q9.

In triangle ABC, right-angled at B, if $\tan A =\frac {1}{\sqrt {3}}$, find the value of: $\text {(i)} \sin A \cos C + \cos A\sin C$ $\text {(ii)} \cos A \cos C + \sin A\sin C$

Ans.

Given that: $triangle ABC$ is a right-angled at B. $\tan A =\frac {1}{\sqrt {3}}$ We know that- $\tan A = \frac {\text {Opposite side of }\angle A}{\text {Adjacent side of }\angle A}=\frac {BC}{AB}=\frac {1}{\sqrt {3}}$ $\text {Let}$$BC = x$ $\text {and}$ $AB = \sqrt{3}x$, $\text {Here }$$x$ $\text {is a positive integer.}$ By Pythagoras theorem in $\triangle ABC$ $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=(\sqrt {3}x)^{2}+ (x)^{2}$ $AC^{2}=3x^{2}+ x^{2}$ $AC^{2}=4x^{2}$ $AC=\sqrt {4x^{2}}$ $AC=2x$ Therefore, $\sin A = \frac {\text {Opposite side to}\angle A}{\text {Hypotenuse}}=\frac {BC}{AC}=\frac {1}{2}$ $\cos A = \frac {\text {Adjacent side to}\angle A}{\text {Hypotenuse}}=\frac {AB}{AC}=\frac {\sqrt{3}}{2}$ $\sin C = \frac {\text {Opposite side to}\angle C}{\text {Hypotenuse}}=\frac {AB}{AC}=\frac {\sqrt{3}}{2}$ $\cos C = \frac {\text {Adjacent side to}\angle C}{\text {Hypotenuse}}=\frac {BC}{AC}=\frac {1}{2}$ $\text {(i)} \sin A \cos C + \cos A\sin C$ By putting the value of the trigonometric function: $=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{4}+\frac {3}{4}=\frac {1+3}{4}=\frac {4}{4}=1$ $\text {(ii)} \cos A \cos C + \sin A\sin C$ By putting the value of the trigonometric function: $=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)=\frac {\sqrt {3}}{4}-\frac {\sqrt {3}}{4}=0$

Q10.

In $\triangle PQR$, right-angled at $Q$, $PR + QR = 25$$\text { cm}$ and $PQ = 5$ $\text { cm}$. Determine the values of $\sin P$, $\cos P$ and $tan P$.

Ans.

Given: In $\triangle PQR$, $\angle Q$ is right-angle. $PQ = 5 \text { cm}$ $PR + QR = 25 \text { cm}$ We have to find $PR$, $QR$ and $\sin P$, $\cos P$ and $\tan P$. Let, In $\triangle PQR$ $PR = x \text { cm}$ Therefore, $PR + QR = 25 \text { cm}$ $x + QR = 25 \text { cm}$ $QR = (25-x) \text { cm}$ By using Pythagoras theorem: $PR^{2}=PQ^{2}+QR^{2}$ Putting value $x^{2}=(5)^{2}+(25-x)^{2}$ by using $(a-b)^{2}=a^{2}-2ab+b^{2}$ $x^{2}=25+625-50x+x^{2}$ $x^{2}-x^{2}+50x=650$ $50x = 650$ $x=\frac {650}{50}=13 \text { cm}$ So, $PR = 13 \text {cm}$ $QR = 25-13 = 12 \text {cm}$ Hence, $\sin P = \frac {\text {Opposite side to}\angle P}{Hypotenuse}=\frac {QR}{PR}=\frac {12}{13}$ $\cos P = \frac {\text {Adjacent side to}\angle P}{Hypotenuse}=\frac {PQ}{PR}=\frac {5}{13}$ $\tan P = \frac {\text {Opposite side to}\angle P}{\text {Adjacent side to}\angle P}=\frac {QR}{PQ}=\frac {12}{5}$

Q11.

State whether the following are true or false. Justify your answer. (i) The value of $tan A$ is always less than $1$. (ii) $\sec A=\frac {12}{5}$ for some value of angle $A$. (iii) $\cos A$ is the abbreviation used for the cosecant of angle $A$. (iv) $\cot A$ is the product of $\cot$ and $A$. (v) $\sin\theta = \frac {4}{3}$,for some angle $\theta$.

Ans.

(i) False, because sides of a right-angled triangle may have any length and also the tangent function $(tan)$ can take values greater than 1. So tan A may have any value. For example : $tan (60^{0})=\sqrt {3}$ which is greater then 1. (ii) True, Because $\sec A$ is the secant function, which is the reciprocal of the cosine function. For example : If $\cos A = \frac {5}{12}$ then $\sec A = \frac {1}{\cos A}=\frac {1}{\frac {5}{12}}=\frac {12}{5}$. Hence $\sec A > 1$. (iii) False, because $\cos A$ represents the cosine function, not the cosecant. The cosecant is represented by $\text {cosec A}$. (iv) False, because $cot A$ is the cotangent function evaluated at $\angle A$. It is not the product of $cot$ and $A$. (v) False, Because the sine function can only take values between $-1$ amd $1$, so $\sin\theta = \frac {4}{3}$ is not possible for any $\angle\theta$.