Exercise 6.4 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.4$
Question 1.
From the top of a tree of height $13 \mathrm{~m}$ the angle of elevation and depression of the top and bottom of another tree are $45^{\circ}$ and $30^{\circ}$ respectively. Find the height of the second tree. $(\sqrt{ } 3=1.732)$
Solution:

In the figure, $A B$ is the 2 nd tree.
$
\begin{aligned}
\tan 45^{\circ} & =\frac{x}{y}=1 \\
x & =y \\
\tan 30^{\circ} & =\frac{1}{\sqrt{3}}=\frac{13}{y} \\
y & =13 \sqrt{3}=x .
\end{aligned}
$
$\therefore$ The height of second tree is $13+x$
$
\begin{aligned}
& =13+13 \sqrt{3} \\
& =13(1+\sqrt{3}) \\
& =13(1+1.732)=13 \times 2.732 \\
& =35.52 \mathrm{~m} .
\end{aligned}
$
Question 2.
A man is standing on the deck of a ship, which is $40 \mathrm{~m}$ above water level. He observes the angle of elevation of the top of a hill as $60^{\circ}$ and the angle of depression of the base of the hill as $30^{\circ}$. Calculate the distance of the hill from the ship and the height of the hill. $(\sqrt{3}=1.732)$
Solution:

$
\begin{aligned}
\tan 60^{\circ} & =\frac{y}{x}=\sqrt{3} \\
y & =\sqrt{3} x \\
\tan 30^{\circ} & =\frac{40}{x}=\frac{1}{\sqrt{3}} \\
x & =40 \sqrt{3} \\
y & =\sqrt{3} x \\
& =\sqrt{3} \times 40 \sqrt{3}=120 \mathrm{~m}
\end{aligned}
$
$\therefore$ The height of the hill $=120+40=160 \mathrm{~m}$
The distance of the hill from the ship is $\mathrm{AC}=\mathrm{x}=40 \sqrt{3} \mathrm{~m}=69.28 \mathrm{~m}$
Question 3.
If the angle of elevation of a cloud from a point ' $h$ ' metres above a lake is $\theta_1$ and the angle of depression of its reflection in the lake is $\theta_2$. Prove that the height that the cloud is located from the ground is $\frac{h\left(\tan \theta_1+\tan \theta_2\right)}{\tan \theta_2-\tan \theta_1}$
Solution:

Let $\mathrm{AB}$ be the surface of the lake and let $\mathrm{p}$ be the point of observation such that $\mathrm{AP}=\mathrm{h}$ meters.

Let $\mathrm{C}$ be the position of the cloud and $\mathrm{C}^{\prime}$ be its reflection in the lake. Then $\mathrm{CB}=\mathrm{C}^{\prime} \mathrm{B}$.
Let $\mathrm{PM}$ be $\perp^{\mathrm{r}}$ from $\mathrm{P}$ on $\mathrm{CB}$
Then $\angle \mathrm{CPM}=\theta_1$, and $\angle \mathrm{MPC}=\theta_2$
Let $\mathrm{CM}=\mathrm{x}$.
Then $\mathrm{CB}=\mathrm{CM}+\mathrm{MB}=\mathrm{CM}+\mathrm{PA}$
$=\mathrm{x}+\mathrm{h}$
In $\Delta$ CPM, we have, $\tan \theta_1=\frac{\mathrm{CM}}{\mathrm{PM}}$
$
\begin{array}{llr}
\Rightarrow \tan \theta_1 & =\frac{x}{\mathrm{AB}} & \\
\Rightarrow \mathrm{AB} & =x \cot \theta_1 & \ldots \ldots(1) \\
& & (\because \mathrm{PM}=\mathrm{AB})
\end{array}
$
In $\Delta \mathrm{PMC}^{\prime}$, we have
$
\begin{array}{ll}
\tan \theta_2 & =\frac{\mathrm{C}^{\prime} \mathrm{M}}{\mathrm{PM}} \\
\Rightarrow \tan \theta_2 & =\frac{x+2 h}{\mathrm{AB}} \\
& \left(\because \mathrm{C}^{\prime} \mathrm{M}=\mathrm{C}^{\prime} \mathrm{B}+\mathrm{BM}=x+h+h\right) \\
\Rightarrow \mathrm{AB} & =(x+2 h) \cot \theta_2
\end{array}
$
From (1)\& (2), we have
$
x \cot \theta_1 \quad=(x+2 h) \cot \theta_2
$
On equating the values of $\mathrm{AB}$.
$
\begin{aligned}
& \Rightarrow x\left(\cot \theta_1-\cot \theta_2\right)=2 h \cot \theta_2 \\
& \Rightarrow x\left(\frac{1}{\tan \theta_1}-\frac{1}{\tan \theta_2}\right)=\frac{2 h}{\tan \theta_2} \\
& \Rightarrow x\left(\frac{\tan \theta_2-\tan \theta_1}{\tan \theta_1 \cdot \tan \theta_2}\right)=\frac{2 h}{\tan \theta_2}
\end{aligned}
$

$
\Rightarrow \quad x=\frac{2 h \tan \theta_1}{\tan \theta_2-\tan \theta_1}
$
Hence the height CB of the cloud is given by
$
\begin{aligned}
& \mathrm{CB}=x+h \\
& \Rightarrow \quad \mathrm{CB}=\frac{2 h \tan \theta_1}{\tan \theta_2-\tan \theta_1}+h \\
& \Rightarrow \quad \mathrm{CB}=\frac{2 h \tan \theta_1+h\left(\tan \theta_2-\tan \theta_1\right)}{\tan \theta_2-\tan \theta_1} \\
& \mathrm{CB}=\frac{h\left(\tan \theta_1+\tan \theta_2\right)}{\tan \theta_2-\tan \theta_1} \text {. } \\
&
\end{aligned}
$
Hence proved
Question 4.
The angle of elevation of the top of a cell phone tower from the foot of a high apartment is $60^{\circ}$ and the angle of depression of the foot of the tower from the top of the apartment is $30^{\circ}$. If the height of the apartment is $50 \mathrm{~m}$, find the height of the cell phone tower. According to radiations control norms, the minimum height of a cell phone tower should be $120 \mathrm{~m}$. State if the height of the above mentioned cell phone tower meets the radiation norms.
Solution:

In the figure $\mathrm{AB}$ is the building $\mathrm{CD}$ is the cell phone tower.
$
\begin{aligned}
\tan 60^{\circ} & =\frac{50+x}{y}=\sqrt{3} \\
50+x & =y \sqrt{3} \\
x & =y \sqrt{3}-50 \\
\tan 30^{\circ} & =\frac{50}{y}=\frac{1}{\sqrt{3}} \\
y & =50 \sqrt{3}
\end{aligned}
$
Substitute $y=50 \sqrt{3}$ in (1)
$
\begin{aligned}
x & =50 \sqrt{3} \times \sqrt{3}-50 \\
& =150-50=100 \mathrm{~m}
\end{aligned}
$
$
\begin{aligned}
\therefore \text { The height of tower } & =50+x=50+100 \\
& =150 \mathrm{~m} .
\end{aligned}
$
Since $150 \mathrm{~m}>120 \mathrm{~m}$, yes the height of the above mentioned tower meet the radiation norms.
Question 5.
The angles of elevation and depression of the top and bottom of a lamp post from the top of a $66 \mathrm{~m}$ high apartment are 600 and $30^{\circ}$ respectively. Find
(i) The height of the lamp post.
(ii) The difference between height of the lamp post and the apartment.
(iii) The distance between the lamp post and the apartment. $(\sqrt{3}=1.732)$
Solution:

Let $\mathrm{AB}$ - Lamp post; $\mathrm{CD}$ - Apartment
In the figure
$
\begin{aligned}
\tan 60^{\circ} & =\frac{x}{y}=\sqrt{3} \\
x & =\sqrt{3} y
\end{aligned}
$

$
\begin{aligned}
& \tan 30^{\circ}=\frac{66}{y}=\frac{1}{\sqrt{3}} \\
& y=66 \sqrt{3} \\
& \therefore \quad x=\sqrt{3} \times 66 \sqrt{3} \\
& \quad \quad(\because y=66 \sqrt{3}) \\
&=66 \times 3=198 \mathrm{~m}
\end{aligned}
$
(i) The height of the Lamp post is $=66+x$
$
\begin{aligned}
& =66+198 \\
\mathrm{AB} & =264 \mathrm{~m}
\end{aligned}
$
(ii) The difference between the height of the Lamp post and the apartment is
$
\begin{aligned}
\mathrm{BM} & =\mathrm{AB}-\mathrm{AM}=264-66 \\
& =198 \mathrm{~m} \quad(\because \mathrm{CD}=\mathrm{AM})
\end{aligned}
$
(iii) The distance between the Lamp post and the apartment
$
y=66 \sqrt{3}
$
$
=114.312 \mathrm{~m}
$
Question 6.
Three villagers $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ can see each $\mathrm{T}$ other across a valley. The horizontal distance between $\mathrm{A}$ and $B$ is $8 \mathrm{~km}$ and the horizontal distance between $B$ and $C$ is $12 \mathrm{~km}$. The angle of depression of $B$ from $\mathrm{A}$ is $20^{\circ}$ and the angle of elevation of $\mathrm{C}$ from $\mathrm{B}$ is $30^{\circ}$. Calculate :

(i) the vertical height between $A$ and $B$.
(ii) the vertical height between $\mathrm{B}$ and $\mathrm{C} .\left(\tan 20^{\circ}=0.3640, \sqrt{3}=1.732\right)$

Solution:

(i) Vertical height between A and B.
$
\begin{aligned}
\tan 20^{\circ} & =\frac{\mathrm{AD}}{8} \\
\mathrm{AD} & =8 \tan 20^{\circ} \\
& \simeq 2.91 \simeq 3 \mathrm{~km}
\end{aligned}
$
(ii) The vertical height between $\mathrm{B}$ and $\mathrm{C}$.
$
\begin{aligned}
& \tan 30^{\circ}=\frac{\mathrm{CE}}{\mathrm{BE}} \\
\mathrm{CE} & =\tan 30 \mathrm{BE} . \\
= & \frac{1}{\sqrt{3}} \times 12=\frac{12 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
= & 4 \sqrt{3} \\
= & 4 \times 1.732 \\
= & 6.93 \mathrm{~km}
\end{aligned}
$