Exercise 6.2 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.2$
Question 1.
Find the angle of elevation of the top of a tower from a point on the ground, which is $30 \mathrm{~m}$ away from the foot of a tower of height $10 \sqrt{3} \mathrm{~m}$.
Solution:

Question 2.
A road is flanked on either side by continuous rows of houses of height $4 \sqrt{3} \mathrm{~m}$ with no space in between them. A pedestrian is standing on the median of the road facing a row house. The angle of elevation from the pedestrian to the top of the house is $30^{\circ}$. Find the width of the road.
Solution:

Question 3.
To a man standing outside his house, the angles of elevation of the top and bottom of a window are $60^{\circ}$ and $45^{\circ}$ respectively. If the height of the man is $180 \mathrm{~cm}$ and if he is $5 \mathrm{~m}$ away from the wall, what is the height of the window? $(\sqrt{3}=1.732)$
Solution:

Let ' $\mathrm{H}$ ' be the fit of the window. Given that elevation of top of the window is $60^{\circ}$.
$
\begin{array}{ll}
\tan 60^{\circ} & =\frac{\mathrm{H}+x}{500}=\sqrt{3} \\
\mathrm{H}+x & =500 \sqrt{3}
\end{array}
$
Given that elevation of bottom of the window is $45^{\circ}$.

$
\begin{aligned}
\therefore \tan 45^{\circ} & =\frac{x}{500}=1 \Rightarrow x=500 \\
\therefore \mathrm{H} & =500 \sqrt{3}-500=866-500 \\
& =366 \mathrm{~cm}=3.66 \mathrm{~m}
\end{aligned}
$
$\therefore$ Height of the window $=3.66 \mathrm{~m}$
Question 4.
A statue $1.6 \mathrm{~m}$ tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $40^{\circ}$. Find the height of the pedestal. ( $\left.\tan 40^{\circ}=0.8391, \sqrt{3}=1.732\right)$

Solution:

Let ' $p$ ' be the fit of the pedestal and $d$ be the distance of statue from point of cabs, on the ground.

Given the elevation of top of the statue from $\mathrm{p}^{\mathrm{f}}$ on ground is $60^{\circ}$.
$
\therefore \tan 60^{\circ}=\frac{1 \cdot 6+p}{d}=\sqrt{3}
$
Also given the elevation of top of the pedestal from point on ground is $40^{\circ}$.
$
\begin{aligned}
\tan 40^{\circ} & =\frac{p}{d}=0.8391 \\
p & =0.8391 d \\
1.6+0.8391 d & =1.732 d \quad 1.6+p=\sqrt{3} d \\
\therefore \quad 0.8929 d & =1.6 \quad 1 \cdot 6+p=1.732 d \\
\therefore \quad d \quad & =1.79 \\
\therefore \text { height of pedestal } & =p=0.839 \times d \\
& =0.839 \times 1.79 \\
& =1.503 \mathrm{~m}
\end{aligned}
$
Question 5.
A flag pole ' $h$ ' metres is on the top of the hemispherical dome of radius $\mathrm{V}$ metres. A man is standing $7 \mathrm{~m}$ away from the dome. Seeing the top of the pole at an angle $45^{\circ}$ and moving $5 \mathrm{~m}$ away from the dome and seeing the bottom of the pole at an angle $30^{\circ}$. Find
(i) the height of the pole
(ii) radius of the dome.

Solution:

Question 6.
The top of a $15 \mathrm{~m}$ high tower makes an angle of elevation of $60^{\circ}$ with the bottom of an electronic pole and angle of elevation of $30^{\circ}$ with the top of the pole. What is the height of the electric pole?

Solution:

Let $\mathrm{BD}$ be tower of height $=15 \mathrm{~m}$ AE be pole of height $=$ ' $p$ '

$
\begin{aligned}
& \text { In } \Delta \mathrm{EBD}, \tan 60^{\circ}=\frac{15}{x}=\sqrt{3} \\
& \therefore \quad x=5 \sqrt{3} \\
& \text { In } \Delta \mathrm{ABC} \text {, } \\
& \tan 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{15-p}{5 \sqrt{3}}=\frac{1}{\sqrt{3}} \\
& \therefore 15-p=5 \\
& p=10 \mathrm{~m} \\
&
\end{aligned}
$
Question 7.
A vertical pole fixed to the ground is divided in the ratio $1: 9$ by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, $25 \mathrm{~m}$ away from the base of the pole, what is the height of the pole?
Solution:

Let $\mathrm{AC}$ be the pole and let point ' $\mathrm{B}$ ' divide it in the ratio $x: 9 x=1: 9$.
Let ' $\mathrm{D}$ ' be the point $25 \mathrm{~m}$.
$
\begin{aligned}
& \tan \alpha \quad=\frac{x}{25} \quad \tan 2 \alpha=\frac{10 x}{25} \\
& \tan 2 \alpha \quad=\frac{2 \tan \alpha}{1-\tan ^2 \alpha} \\
& \frac{10 x^5}{25}=\frac{2 \times \frac{x}{25}}{1-\frac{x^2}{625}} \\
& 5-\frac{5 x^2}{625}=1 \Rightarrow \frac{5 x^2}{625}=4 \Rightarrow x^2=\frac{4 \times 625^{125}}{8}=500 \\
& \text { Height of pole }=10 x \\
& x=100 \sqrt{5} \mathrm{~m}
\end{aligned}
$
Question 8.
A traveler approaches a mountain on highway. He measures the angle of elevation to the peak at each milestone. At two consecutive milestones the angles measured are $4^{\circ}$ and $8^{\circ}$. What is the height of the peak if the distance between consecutive milestones is 1 mile, $\left(\tan 4^{\circ}=0.0699\right.$, $\tan 8^{\circ}$ $=0.1405)$.
Solution:

Let $\mathrm{AB}$ denote the height of the peak and be ' $h$ '.
In $\triangle \mathrm{ABC}$,

$
\begin{array}{rlr}
\tan 8^{\circ} & =\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{h}{m} \\
m & =\frac{h}{\tan 8}
\end{array}
$
In $\triangle \mathrm{ABD}$,
$
\begin{aligned}
\tan 4^{\circ} & =\frac{\mathrm{AB}}{\mathrm{BD}}=\frac{h}{m+1} \\
m+1 & =\frac{h}{\tan 4}
\end{aligned}
$
From (1) and (2)
$
\begin{aligned}
& \quad \frac{h}{\tan 8}+1=\frac{h}{\tan 4} \\
& 1=\frac{h}{\tan 4}-\frac{h}{\tan 8} \\
& h\left[\frac{\tan 8-\tan 4}{\tan 4 \tan 8}\right]=1 \\
& h=\frac{\tan 4 \times \tan 8}{\tan 8-\tan 4} \\
& =\frac{.0699 \times .1405}{.1405-.0699} \\
& =\frac{.00982}{.0706}=0.14 \text { mile (approx) }
\end{aligned}
$