Exercise 6.1 - Chapter 6 Trigonometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex 6.1 : Chapter 6 - Trigonometry - 9th Maths Guide Samacheer Kalvi Solutions
Question $1 .$
From the given figure find all the trigonometric ratios of angle B.
Solution:
$\sin B=\frac{9}{41}$
$\cos B_{i}=\frac{40}{41}$
$\tan \mathrm{B}=\frac{9}{40}$
$\operatorname{cosec} \mathrm{B}=\frac{1}{\sin \mathrm{B}}=\frac{41}{9}$
$\sec B=\frac{1}{\cos B}=\frac{41}{40}$
$\cot B=\frac{1}{\tan B}=\frac{40}{9}$

Question 2.
From the given figure, find the values of
(i) $\sin B$
(ii) $\sec B$
(iii) $\cot \mathrm{B}$
(iv) $\cos C$
(v) $\tan C$
(vi) $\operatorname{cosec} \mathrm{C}$
Solution:

From the figure

(i) $\quad \sin \mathrm{B}=\frac{12}{13}$
(ii)$\sec B=\frac{1}{\cos B} \quad$ B $5 \mathrm{D}$
$=\frac{1}{\frac{5}{13}}=\frac{13}{5}$
(iii) $\quad \cot \mathrm{B}=\frac{1}{\tan \mathrm{B}}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$
(iv) $\quad \cos \mathrm{C}=\frac{\frac{4}{16}}{2 \underset{5}{20}}=\frac{4}{5}$
(v) $\tan \mathrm{C}=\frac{\frac{3}{12}_{16}^{4}=\frac{3}{4}}{}$
(vi) $\operatorname{cosec} \mathrm{C}=\frac{1}{\sin \mathrm{C}}=\frac{12}{20}=\frac{20}{12}=\frac{5}{3}$

Question 3 .
If $2 \cos \theta=\sqrt{3}$, then find all the trigonometric ratios of angle $\theta$.

Solution:
$\text { If } \begin{aligned}
2 \cos \theta &=\sqrt{3} \\
\cos \theta &=\frac{\sqrt{3}}{2}
\end{aligned}$
By the Pythagoras theorem,
$\begin{aligned}
x=\sqrt{2^{2}-\sqrt{3}^{2}}=\sqrt{4-3} &=\sqrt{1}=1 \\
\therefore \sin \theta &=\frac{1}{2} \\
\cos \theta &=\frac{\sqrt{3}}{2} \\
\tan \theta &=\frac{1}{\sqrt{3}} \\
\operatorname{cosec} \theta &=2 \\
\sec \theta &=\frac{2}{\sqrt{3}} \\
\cot \theta &=\sqrt{3}
\end{aligned}$

Question 4 .

If $\cos A=\frac{3}{5}$, then find the value of $\frac{\sin A-\cos A}{2 \tan A}$

Solution:
$\begin{aligned}
\sin A &=\frac{4}{5} \\
\tan A &=\frac{4}{3}
\end{aligned}$

$\therefore \frac{\sin A-\cos A}{2 \tan A}=\frac{\frac{4}{5}-\frac{3}{5}}{2 \times \frac{4}{3}}=\frac{\frac{1}{5}}{2 \times \frac{4}{3}}=\frac{1}{2} \times \frac{1}{5} \times \frac{3}{4}=\frac{3}{40}$

Question 5 .
If $\cos A=\frac{2 x}{1+z^{2}}$ then find the values of $\sin A$ and $\tan A$ in terms of $x$.

Solution:
By the pythagoras theorem, $\mathrm{AB}^{2}=\mathrm{OA}^{2}+\mathrm{OB}^{2}$
$\left(1+x^{2}\right)^{2}=(2 x)^{2}+\mathrm{OB}^{2}$
$\begin{aligned}
\mathrm{OB}^{2}=\left(1+x^{2}\right)^{2}-(2 x)^{2} &=1+x^{4}+2 x^{2}-4 x^{2} \\
&=1+x^{4}-2 x^{2} \\
\mathrm{OB}^{2} &=\left(1-x^{2}\right)^{2} \\
\mathrm{OB} &=\left(1-x^{2}\right) \\
\therefore \sin \mathrm{A} &=\frac{1-x^{2}}{1+x^{2}} \\
\tan \mathrm{A} &=\frac{1-x^{2}}{2 x}
\end{aligned}$

Question 6.
If $\sin \theta=\frac{a}{\sqrt{a^{2}+b^{2}}}$, then show that $b \sin \theta=a \cos \theta$.
Solution:
$\begin{aligned}
\sin \theta &=\frac{a}{\sqrt{a^{2}+b^{2}}} \\
\cos \theta &=\frac{b}{\sqrt{a^{2}+b^{2}}} \\
b \sin \theta &=b \times \frac{a}{\sqrt{a^{2}+b^{2}}}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \ldots(1)
\end{aligned}$

$a \cos \theta=a \times \frac{b}{\sqrt{a^{2}+b^{2}}}=\frac{a b}{\sqrt{a^{2}+b^{2}}} \ldots(2)$
$(1)=(2) \Rightarrow$ LHS $=$ R.H.S.
Hence proved.

Question 7.
If $3 \cot A=2$, then find the value of $\frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}$

Solution:
$3 \cot \mathrm{A}=2$
$\cot \mathrm{A}=\frac{2}{3}$
$\cot \mathrm{A}=\frac{\text { Adjacent side }}{\text { Opposite side }}$
$\tan \mathrm{A}=\frac{\text { Opp. side }}{\text { Adj. side }}=\frac{3}{2}$
$\sin A=\frac{3}{\sqrt{13}}$
$\cos A=\frac{2}{\sqrt{13}}$
$\therefore \frac{4 \sin A-3 \cos A}{2 \sin A+3 \cos A}=\frac{4 \times \frac{3}{\sqrt{13}}-3 \times \frac{2}{\sqrt{13}}}{2 \times \frac{3}{\sqrt{13}}+3 \times \frac{2}{\sqrt{13}}}$
$=\frac{\frac{12}{\sqrt{13}}-\frac{6}{\sqrt{13}}}{\frac{6}{\sqrt{13}}+\frac{6}{\sqrt{13}}}=\frac{\frac{6}{\sqrt{13}}}{\frac{12}{\sqrt{13}}}=\frac{6}{\sqrt{13}} \times \frac{\sqrt{13}}{12}=\frac{6}{12}=\frac{1}{2}$

Question $8 .$
If $\cos \theta: \sin \theta=1: 2$, then find the value of $\frac{8 \cos \theta-2 \sin \theta}{4 \cos \theta+2 \sin \theta}$
Solution:
$\begin{aligned}
\cos \theta: \sin \theta &=1: 2 \\
\frac{\cos \theta}{\sin \theta} &=\frac{1}{2} \\
\therefore \cos \theta &=\frac{1}{2} \sin \theta \\
\sin \theta &=2 \cos \theta
\end{aligned}$

$\therefore \frac{8 \cos \theta-2 \sin \theta}{4 \cos \theta+2 \sin \theta}=\frac{48 \times \frac{1}{2} \sin \theta-2 \sin \theta}{\frac{2}{4 \times \frac{1}{8} \sin \theta+2 \sin \theta}}=\frac{4 \sin \theta-2 \sin \theta}{2 \sin \theta+2 \sin \theta}=\frac{2 \sin \theta}{4 \sin \theta}=\frac{1}{2}$
$\therefore \frac{8 \cos \theta-2 \sin \theta}{4 \cos \theta+2 \sin \theta}=\frac{1}{2}$
 

Question $9 .$
From the given figure, prove that $\theta+\phi=90^{\circ}$. Also prove that there are two other right angled triangles.

Find $\sin \alpha, \cos \beta$ and $\tan \phi$
Solution:
In $\triangle \mathrm{ABC}$
$\begin{aligned}
A C^{2} &=15^{2}=225 \\
B C^{2} &=20^{2}=400 \\
\mathrm{AB}^{2} &=(9+16)^{2} \\
&=25^{2}=625
\end{aligned}$
From (1), (2), (3)
$\begin{aligned}
A B^{2} &=A C^{2}+B C^{2} \\
625 &=225+400=625 \\
\therefore \angle C &=\theta+\phi=90^{\circ}
\end{aligned}$

( $\therefore$ By Pythagoras theorem, in a right angled triangle square of hypotenuse is equal to sum of the squares of other two side)
And also in the figure, $\triangle \mathrm{ADC}, \triangle \mathrm{DBC}$ are two other triangles.
As per the data given,
$9^{2}+12^{2}=81+144=225=15^{2}$
$\therefore \triangle \mathrm{ADC}$ is a right angled triangle, then $12^{2}+16^{2}=144+256=400=20^{2}$
$\therefore \triangle \mathrm{DBC}$ is also a right angled triangle
$\sin \alpha=\frac{12}{15}=\frac{4}{5}, \cos \beta=\frac{16}{20}=\frac{4}{5}, \tan \phi=\frac{16}{12}=\frac{4}{3}$
 

Question $10 .$
A boy standing at point $\mathrm{O}$ finds his kite flying at a point $\mathrm{P}$ with distance $\mathrm{OP}=25 \mathrm{~m}$. It is at a height of $5 \mathrm{~m}$ from the ground. When the thread is extended by $10 \mathrm{~m}$ from $\mathrm{P}$, it reaches a point $\mathrm{Q}$. What will be the height QN of the kite from the ground? (use trigonometric ratios)
Solution:
In the figure,

$\triangle \mathrm{OPM}, \triangle \mathrm{OQN}$ are similar triangles. In similar triangles the sides are in the same proportional.

$\begin{aligned}
\therefore \frac{\mathrm{QN}}{\mathrm{PM}} &=\frac{\mathrm{QO}}{\mathrm{PO}} \\
\frac{h}{5} &=\frac{35}{25} \\
h &=\frac{5 \times 35}{25} \\
h &=\frac{8 \times 35^{7}}{25} \not{\beta} \\
h &=7 \mathrm{~m} .
\end{aligned}$