Exercise 6.3 - Chapter 6 Trigonometry 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $6.3$
Question 1.
From the top of a rock $50 \sqrt{3} \mathrm{~m}$ high, the angle of depression of a car on the ground is observed to be $30^{\circ}$. Find the distance of the car from the rock.
Solution:

In the figure
$
\begin{aligned}
\tan 30 & =\frac{\mathrm{AB}}{\mathrm{BC}} \\
\frac{1}{\sqrt{3}} & =\frac{50 \sqrt{3}}{d} \\
d & =50 \sqrt{3} \times \sqrt{3} \\
& =50 \times 3=150 \mathrm{~m} .
\end{aligned}
$
Question 2.
The horizontal distance between two buildings is $70 \mathrm{~m}$. The angle of depression of the top of the first building when seen from the top of the second building is $45^{\circ}$. If the height of the second building is $120 \mathrm{~m}$, find the height of the first building.
Solution:

In the figure, $\tan \angle \mathrm{C}=\tan 45^{\circ}=1$.
$
\begin{aligned}
1 & =\frac{\mathrm{AB}}{70}=\frac{x}{70} \\
x & =70 \mathrm{~m} . \\
\therefore \quad \mathrm{BD} & =120-\mathrm{AB} \\
h & =120-70=50 \mathrm{~m} .
\end{aligned}
$
$\therefore$ The height of the first building is $50 \mathrm{~m}$.
Question 3.
From the top of the tower $60 \mathrm{~m}$ high the angles of depression of the top and bottom of a vertical lamp post are observed to be $38^{\circ}$ and $60^{\circ}$ respectively. Find the height of the lamp post, $\left(\tan 38^{\circ}=\right.$ $0.7813, \sqrt{3}=1.732)$
Solution:

From the figure,

$
\begin{aligned}
\tan \angle \mathrm{E} & =\tan 60^{\circ} \\
\sqrt{3} & =\frac{60}{y} \\
\sqrt{3} y & =60 \\
y & =\frac{60 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
y & =\frac{60 \sqrt{3}}{3}=20 \sqrt{3} \\
\tan \angle \mathrm{C} & =\frac{x}{y} \\
0.7813 & =\frac{x}{20 \sqrt{3}} \\
x & =20 \sqrt{3} \times 0.7813 \\
& =20 \times 1.732 \times 0.7813 \\
& =27.064
\end{aligned}
$
$\therefore$ The height of the lamp post $=\mathrm{CE}$
$
\mathrm{CE}=\mathrm{BD}=60-27.064=32.93 \mathrm{~m} \text {. }
$
Question 4.
An aeroplane at an altitude of $1800 \mathrm{~m}$ finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are $60^{\circ}$ and $30^{\circ}$ respectively. Find the distance between the two boats. $(\sqrt{3}=1.732)$
Solution:

In the figure
$
\begin{aligned}
& \tan \mathrm{C}=\tan 60^{\circ}=\sqrt{3} \\
& \tan \mathrm{D}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}
\end{aligned}
$
$(1),(2)$ gives, $\tan 60^{\circ}=\sqrt{3}$,
$
\begin{aligned}
\frac{1800}{x} & =\sqrt{3} \\
\sqrt{3} x & =1800 \\
x & =\frac{1800 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} \\
& =\frac{1800 \sqrt{3}}{3}=600 \sqrt{3} \\
\tan 30^{\circ} & =\frac{1}{\sqrt{3}}=\frac{1800}{d+x} \\
d+x & =1800 \sqrt{3} \\
d+600 \sqrt{3} & =1800 \sqrt{3} \\
d & =1800 \sqrt{3}-600 \sqrt{3} \\
& =1200 \sqrt{3} .
\end{aligned}
$
Distance between the boats $=1200 \sqrt{3} \mathrm{~m}$
$
=2078.4 \mathrm{~m}
$
Question 5.
From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be $30^{\circ}$ and $60^{\circ}$. If the height of the lighthouse is $\mathrm{h}$ meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is $\frac{4 h}{\sqrt{3}} \mathrm{~m}$.
Solution:

$\therefore$ The distance between the ships $\mathrm{A} \& \mathrm{~B}=x+y$
$
\begin{aligned}
& =\sqrt{3} h+\frac{h}{\sqrt{3}} \\
=\frac{3 h+h}{\sqrt{3}} & =\frac{4 h}{\sqrt{3}} \mathrm{~m} .
\end{aligned}
$
It is proved.
Question 6.
A lift in a building of height 90 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is $60^{\circ}$. Two minutes later, the angle of depression reduces to $30^{\circ}$. If the fountain is $30 \sqrt{3}$ feet from the entrance of the lift, find the speed of the lift which is descending.
Solution:

$
\begin{aligned}
\therefore \quad \mathrm{AB} & =\text { speed } \times 2 \quad(\mathrm{D}=s t) \\
\text { In } \Delta \mathrm{AED}, \tan 60^{\circ} & =\frac{\mathrm{DE}}{\mathrm{AE}} \\
=\frac{\mathrm{DE}}{30 \sqrt{3}} & =\sqrt{3} \Rightarrow \mathrm{DE}=90 \mathrm{ft.}
\end{aligned}
$
In $\Delta \mathrm{BFD}, \tan 30^{\circ}=\frac{\mathrm{DF}}{\mathrm{BF}}=\frac{1}{\sqrt{3}}$
$
\begin{aligned}
\frac{\mathrm{DF}}{30 \sqrt{3}}=\frac{1}{\sqrt{3}} & \Rightarrow \mathrm{DF}=30 \mathrm{ft} \\
\therefore \mathrm{EF} & =\mathrm{DE}-\mathrm{DF}=90-30=60 \mathrm{ft} . \\
\therefore \mathrm{AB} & =\mathrm{EF}=60 \mathrm{ft} \\
\text { Speed } & =\frac{\text { Distance }}{\text { time }}=\frac{60}{2} \mathrm{ft} / \mathrm{min} \\
& =30 \mathrm{ft} / \mathrm{min} \\
& =\frac{30 \times .305}{60} \mathrm{~m} / \mathrm{sec} \\
& =.15 \mathrm{~m} / \mathrm{sec}[\because 1 \mathrm{foot}=.305 \mathrm{~m}]
\end{aligned}
$