# 29. Height and Distance Solutions Part 2¶

The diagram is given below:

Let $AB$ be the lamp post having height $h$ m, and $BD$ be the girl having height $1.6$ m. The distance of the grl from the lamp post is $AC = 3.2$ m. $CE$ is the langeth of the shadow given as $4.8$ m. In the $\triangle ABE$ and $\triangle CDE, \angle E$ is common, $\angle A = \angle C = 90^\circ$ so third angle will be also equal. This makes the triangles similar.

$\therefore \frac{AB}{CD} = \frac{AE}{CE} \Rightarrow h = \frac{8}{3}$ m.

The diagram is given below:

Let $AC$ be the building having a height of $30$ m. Let $E$ and $G$ point of observations where angles of elevation are $60^\circ$ and $30^\circ$ respectively. Let $AEF$ be the line of foot of the building and foot of the observer which is a horizontal line. Let $DE$ and $FE$ are the heights of the observer. Draw $BEG\parallel ADF$ so that $AB = DE = FG = 1.5$ m. Thus, $BC = 28.5$ m. We have to find $DF = EG$.

In $\triangle BCE, \tan60^\circ = \sqrt{3} = \frac{BC}{CE} \Rightarrow CE = \frac{28.5}{\sqrt{3}}$ m.

In $\triangle BCG, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{CG} \Rightarrow CG = 28.5\sqrt{3}$ m.

Thus, $DF = EG = CG - CF = \frac{57}{\sqrt{3}}$ m, which is the distance walked by the observer.

The diagram is given below:

Let the height of the tower $AB$ is $h$ m. When the altitude of the sun is $60^\circ$ let the length of the shadown be $AC = x$ m. Then according to question length of shadow when the sun’s altitude i $30^\circ$ the length of shadow will be $AD, 40$ m longer i.e. $AD = x + 40$.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{h}{x} \Rightarrow x = \sqrt{3}h$ m.

In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AC} = \frac{h}{x + 40} \Rightarrow x = 20$ m and $h = 20\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the building with $20$ m height. Let the height of tower be $h$ m represented by $BC$ in the figure. Let $D$ be the point of observation at a distance $x$ from the foot of the building $AB$.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{20}{x} \Rightarrow x = 20\sqrt{3}$ m.

In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{AD} = \frac{h + 20}{x}\Rightarrow h = 40$ m.

The diagram is given below:

Let $DE$ be the building having a height of $8$ m. Let $AC$ be the multistoried building having height $h + 8$ m. Foot of both the buildings are joined on horizontal plane i.e. $AD$. Draw a line parallel to $AD$ which is $BE$. So $BE$ is equal to $AD$ which we have let as $x$ m. Clearly, $AB = 8$ m. Let height of $BC$ to be $h$ m.

In $\triangle CBE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BC}{BE} = \frac{h}{x} \Rightarrow \sqrt{3}h = x$.

In $\triangle ACD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AC}{AD} = \frac{h + 8}{x}\Rightarrow h = \frac{8}{\sqrt{3} - 1} \Rightarrow h + 8 = \frac{8\sqrt{3}}{\sqrt{3} - 1}$ m.

The diagram is given below:

Let $AB$ be the pedestal having height $h$ m and $BC$ be the statue having height $1.6$ m on top of pedestal. Let $D$ be the point of observation from where the angles of elevation as given in the question are $45^\circ$ and $60^\circ$.

In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{BD} = \frac{h}{x} \Rightarrow h = x$.

In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{CD} = \frac{h + 1.6}{x} \Rightarrow h = \frac{1.6}{\sqrt{3} - 1}$ m.

This problem is similar to 55 and has been left as an exercise.

The diagram is given below:

Let $AB$ be the tower having height $75$ m. Let $C$ and $D$ be the position of two ships and angles of elevation are as given in the question. Let foor of the tower be in line with ships such that $AC = x$ m and distance between the ships as $d$ m.

In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC} = \frac{75}{x} \Rightarrow x = 75$ m.

In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{75}{x + d} \Rightarrow d = 75(\sqrt{3} - 1)$ m.

The diagram is given below:

Let $AB$ be the building and $CD$ be thw tower having height $50$ m. The angles of elevation are shown as given in the question. Let distance between the foot of the tower and the building be $d$ m and height of the building be $h$ m.

In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{d} \Rightarrow d = \sqrt{3}h$.

In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{50}{d} \Rightarrow 3h = 50 \Rightarrow h = \frac{50}{3}$ m.

The diagram is given below:

Let $DE$ represent the banks of river and $BC$ the bridge. Given that height of the bridge is $30$ m. $\therefore BD = CE = 30$ m. The angles of depression from point $A$ is shown as given in the question. We have to find $DE = BC$ i.e. width of the river.

In $\triangle ACE, \tan45^\circ = 1 = \frac{CE}{AC} \Rightarrow AC = 30$ m.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BD}{AB} \Rightarrow AB = 30\sqrt{3}$ m.

Thus, width of river $= 30 + 30\sqrt{3} = 30(\sqrt{3} + 1)$ m

The diagram is given below:

Let $BC$ and $DE$ be the poles, both having the same height $h$ m. Let $BD$ be the road having a width of $80$ m. Also, let $A$ be the point of observation from which angles of observation are shown as given in the question. Let $AB = x$ m and thus $AD = 80 - x$ m.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AB} \Rightarrow BC = h = x\sqrt{3}$ m.

In $\triangle ADE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{DE}{AD} \Rightarrow DE = h = \frac{80 - x}{\sqrt{3}}$ m.

Since the poles have equal height, therefore $\sqrt{3}x = \frac{80 - x}{\sqrt{3}}\Rightarrow x = 20$ m.

$\therefore h = 20\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the tree having height $20$ m, $CD$ the width of river. Let $C$ and $D$ be the foot of the poles on opposite banks of the river such that foot of the poles and the tree are in the same horizontal plane. Let $AC = x$ and $AD = y$ m. The angles of depression are shown as given in the question.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} \Rightarrow AC = \frac{20}{\sqrt{3}}$ m.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} \Rightarrow AD = 20\sqrt{3}$ m.

Thus, $CD = AC + AD = \frac{80}{\sqrt{3}}$ m.

The diagram is given below:

Let $AB$ be the tower having height $h$ m, $BC$ be the flag staff having height $7$ m. Also, let point of observation at $D$ having angles of observation as $30^\circ$ and $45^\circ$ as given in the question such that $AD = x$ m.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{x} \Rightarrow \sqrt{3}h = x$.

In $\triangle ACD, \tan45^\circ = 1 = \frac{AC}{AD} = \frac{h + 7}{x} \Rightarrow h = \frac{7}{\sqrt{3} - 1}$ m.

The diagram is given below:

Let $AC$ and $AD$ be the length of shadows with angles of depression $45^\circ$ and $30^\circ$ respectively such that $CD = 2x$ m. Also let $AB$ as the tower having height $h$ m and $AC$ as the length of shadow with angle of depression $45^\circ$ having length $d$ m.

In $\triangle ABC, \tan45^\circ = 1 = \frac{AB}{AC} = \frac{h}{d} \Rightarrow h = d$ m.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{2x + d} \Rightarrow h = \frac{2x}{\sqrt{3} - 1} = \frac{2x(\sqrt{3} + 1)}{3 - 1} = x(\sqrt{3} + 1)$ m.