Samacheer Kalvi 10th Maths Exercise 6.4 Solutions

Let $AB$ be the first tree of height $13$ m and $CD$ the second tree.

From the top $A$, the angle of depression of the foot $D$ is $30^\circ$ and the angle of elevation of the top $C$ is $45^\circ$.

Let the distance between the trees be $d$ and let $CE = x$, where $E$ is the point on $CD$ at the level of $A$.

In the right triangle with the $30^\circ$ angle, $\tan 30^\circ = \dfrac{13}{d}$

$\dfrac{1}{\sqrt{3}} = \dfrac{13}{d} \Rightarrow d = 13\sqrt{3}$ m

In the right triangle with the $45^\circ$ angle, $\tan 45^\circ = \dfrac{x}{d}$

$1 = \dfrac{x}{13\sqrt{3}} \Rightarrow x = 13\sqrt{3}$ m

Height of second tree $= 13 + 13\sqrt{3} = 13(1+\sqrt{3})$

$= 13(1+1.732) = 13 \times 2.732 = 35.516$ m

Let $A$ be the man's position on the deck, $40$ m above the water level $B$.

Let $CD$ be the hill, $D$ its base, and let $BD = d$ be the horizontal distance.

Angle of depression of the base $D$ is $30^\circ$, so $\tan 30^\circ = \dfrac{40}{d}$

$\dfrac{1}{\sqrt{3}} = \dfrac{40}{d} \Rightarrow d = 40\sqrt{3} = 40 \times 1.732 = 69.28$ m

Let $x$ be the height of the hill above the level of the deck.

$\tan 60^\circ = \dfrac{x}{d} \Rightarrow \sqrt{3} = \dfrac{x}{40\sqrt{3}}$

$x = 40\sqrt{3} \times \sqrt{3} = 120$ m

Height of the hill $= 120 + 40 = 160$ m

Let $P$ be the point $h$ m above the lake and let the cloud $C$ be at height $H$ m above the lake.

The reflection $C'$ of the cloud lies $H$ m below the lake surface.

Let the horizontal distance from $P$ to the vertical line through the cloud be $x$.

Height of the cloud above the level of $P$ is $H - h$, so $\tan\theta_1 = \dfrac{H-h}{x}$ ...(1)

Depth of the reflection below the level of $P$ is $H + h$, so $\tan\theta_2 = \dfrac{H+h}{x}$ ...(2)

From (1), $x = \dfrac{H-h}{\tan\theta_1}$ and from (2), $x = \dfrac{H+h}{\tan\theta_2}$

Hence $(H-h)\tan\theta_2 = (H+h)\tan\theta_1$

$H\tan\theta_2 - h\tan\theta_2 = H\tan\theta_1 + h\tan\theta_1$

$H(\tan\theta_2 - \tan\theta_1) = h(\tan\theta_1 + \tan\theta_2)$

$H = \dfrac{h(\tan\theta_1 + \tan\theta_2)}{\tan\theta_2 - \tan\theta_1}$

Let $AB$ be the apartment of height $50$ m and $CD$ the cell phone tower, with $BD = d$ the distance between them.

From the top $A$ of the apartment, the angle of depression of the foot $D$ of the tower is $30^\circ$.

$\tan 30^\circ = \dfrac{50}{d} \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{50}{d}$

$d = 50\sqrt{3}$ m

From the foot $B$ of the apartment, the angle of elevation of the top $C$ of the tower is $60^\circ$.

$\tan 60^\circ = \dfrac{CD}{d} \Rightarrow \sqrt{3} = \dfrac{CD}{50\sqrt{3}}$

$CD = 50\sqrt{3} \times \sqrt{3} = 150$ m

Since $150 > 120$, the tower satisfies the radiation norms.

Let $AB$ be the apartment of height $66$ m and $CD$ the lamp post, with distance $BD = d$.

From the top $A$, the angle of depression of the foot $D$ is $30^\circ$.

$\tan 30^\circ = \dfrac{66}{d} \Rightarrow d = 66\sqrt{3}$ m

Let $x$ be the part of the lamp post above the level of $A$.

$\tan 60^\circ = \dfrac{x}{d} \Rightarrow \sqrt{3} = \dfrac{x}{66\sqrt{3}}$

$x = 66\sqrt{3} \times \sqrt{3} = 198$ m

Height of the lamp post $= 198 + 66 = 264$ m