Exercise 6.4 - Chapter 6 Trigonometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.4$
Question $1 .$
Find the value of the following:
(i) $\sin 49^{\circ}$
(ii) $\cos 74^{\circ} 39^{\prime}$
(iii) $\tan 54^{\circ} 26^{\prime}$
(iv) $\sin 21^{\circ} 21^{\prime}$
(v) $\cos 33^{\circ} 53^{\prime}$
(vi) $\tan 70^{\circ} 17^{\prime}$
Solution:

$\sin 49^{\circ}=0.7547
$(ii) $\cos 74^{\circ} 39^{\prime}$
$\cos 74^{\circ} 36^{\prime}=0.2656$ (From the natural cosines table)
Mean difference $3^{\prime}=\frac{8(-)}{0.2648}$$
\cos 74^{\circ} 39^{\prime}=\overline{0.2648}$
(iii) $\tan 54^{\circ} 26^{\prime}$
From the natural tangents table
$\begin{aligned}
\tan 54^{\circ} 24^{\prime} &=1.3968 \\
\text { Mean difference } 2^{\prime} &=\frac{17}{1.3985} \\
\tan 54^{\circ} 26^{\prime} &=\underline{1 .}
\end{aligned}$

Question $2 .$
Find the value of $\theta$ if
(i) $\sin \theta=0.9975$
(ii) $\cos \theta=0.6763$
(iii) $\tan \theta=0.0720$
(iv) $\cos \theta=0.0410$
(v) $\tan \theta=7.5958$
Solution:
(i) From the natural sines table
$\begin{aligned}
\sin 85^{\circ} 57^{\prime} &=0.9975 \\
\therefore \theta &=85^{\circ} 57^{\prime} \\
\cos \theta &=0.6763 \\
\cos 47^{\circ} 27^{\prime} &=0.6763 \\
\therefore \theta &=47^{\circ} 27^{\prime}
\end{aligned}$
(ii)$\begin{aligned} \cos \theta &=0.6763 \\ \cos 47^{\circ} 27^{\prime} &=0.6763 \\ \therefore \theta &=47^{\circ} 27^{\prime} \end{aligned}$
(iii)
$\begin{aligned}
\tan \theta &=0.0720 \\
\tan 4^{\circ} 7^{\prime} &=0.0720 \\
\therefore \theta &=4^{\circ} 7^{\prime}
\end{aligned}$
(iv) $\cos \theta=0.0410$
$\begin{aligned}
\cos 87^{\circ} 39^{\circ} &=0.0410 \\
\therefore \theta &=87^{\circ} 39^{\circ}
\end{aligned}$
(v) $\tan \theta=7.5958$
$\begin{aligned}
\tan 82^{\circ} 30^{\prime} &=7.5958 \\
\therefore \theta &=82^{\circ} 30^{\prime}
\end{aligned}$

Question $3 .$
Find the value of the following :

(i) $\sin 65^{\circ} 39^{\prime}+\cos 24^{\circ} 57^{\prime}$
(ii) $\tan 70^{\circ} 58^{\prime}+\cos 15^{\circ} 26^{\prime}-\sin 84^{\circ} 59^{\prime}$
Solution:
(i) $=0.9111+0.9066+0.1793=1.9970$
(ii) $=2.8982+0.9639-0.9962=3.8621-0.9962=2.8659$

Question $4 .$
Find the area of a right triangle whose hypotenuse is $10 \mathrm{~cm}$ and one of the acute angle is $24^{\circ} 24^{\prime}$
Solution:

$\begin{aligned}
\text { Hypotenuse }=10 \mathrm{~cm} & \\
\text { One of the acute angle } &=24^{\circ} 24^{\prime} \\
\sin 24^{\circ} 24^{\prime} &=0.4131 \\
\frac{x}{10} &=0.4131 \\
x &=0.4131 \times 10 \quad x \\
x &=4.131 \\
\cos 24^{\circ} 24^{\prime} &=0.9107 \\
\frac{y}{10} &=0.9107 \\
y &=9.107 \\
\therefore \text { Area of the triangle } &=\frac{1}{2} b h \\
&=\frac{1}{2} \times y \times x=\frac{1}{2} \times 9.107 \times 4.131=18.81 \mathrm{sq} . \mathrm{cm} .
\end{aligned}$

Question $5 .$
Find the angle made by a ladder of length $5 \mathrm{~m}$ with the ground, if one of its end is $4 \mathrm{~m}$ away from the wall and the other end is on the wall.
Solution:

$\begin{aligned}
\cos \theta &=\frac{4}{5}=0.8 \\
\cos 36^{\circ} 52^{\prime} &=0.8 \\
\therefore \theta &=36^{\circ} 52^{\prime}
\end{aligned}$

Question $6 .$
In the given figure, HT shows the height of a tree standing vertically. From a point $P$, the angle of elevation of the top of the tree (that is $\angle \mathrm{P}$ ) measures $42^{\circ}$ and the distance to the tree is 60 metres. Find the height of the tree.

Solution:
$\begin{aligned}
\tan 42^{\circ} &=\frac{h}{60}=0.9004 \\
h &=0.9004 \times 60=54.024 \mathrm{~m}
\end{aligned}$