# 28. Height and Distance Solutions Part 1¶

The diagram is given below:

Let $BC$ be the tower, $A$ the point of observation and $\theta$ as angle of elevation.

Since the tower is vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\tan\theta = \frac{BC}{AB} = \frac{100\sqrt{3}}{100} = \sqrt{3} \Rightarrow \theta = 60^\circ$.

The diagram is given below:

Let $BC$ be the tower, $A$ the point of observation and the angle of elevation is $30^\circ$.

Since the tower is vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\tan30^\circ = \frac{BC}{AB} \Rightarrow BC = 30.\frac{1}{\sqrt{3}} = 10\sqrt{3}$ m.

The diagram is given below:

Let $BC$ be the height of kite, $AC$ be the length of string the angle of elevation is $60^\circ$.

Since the kite would be vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\sin60^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{60.2}{\sqrt{3}} = 40\sqrt{3}$ m.

The diagram is given below:

Let $BC$ be the height of kite, $AC$ be the length of string the angle of elevation is $60^\circ$.

Since the kite would be vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\sin60^\circ = \frac{BC}{AC} \Rightarrow BC = 100\frac{\sqrt{3}}{2}= 50\sqrt{3}$ m.

The diagram is given below:

Let $BC$ be the pole, $A$ the point where rope is tied to the ground and the angle of elevation is $30^\circ$.

Since the pole is vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\sin30^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{12}{\sin30^\circ} = 24$ m.

Thus, the acrobat has to climb $24$ m.

The diagram is given below:

Let $BC$ be the pole, $A$ the point where rope is tied to the ground and the angle of elevation is $30^\circ$.

Since the pole is vertical it forms a right-angle triangle with right angle at $B$. Thus,

$\sin30^\circ = \frac{BC}{AC} \Rightarrow BC = 20.\frac{1}{2} = 10$ m.

The diagram is given below:

Let the shaded region represent the river and vertical lines the banks. $AB$ represents the bridge, making an angle of $45^\circ$ with the bank. Let $BC$ represent the width of river, which clearly makes a right angle triangle with right angle at $C$.

Clearly, $\sin45^\circ = \frac{BC}{AB} \Rightarrow BC = 150.\frac{1}{\sqrt{2}} = 75\sqrt{2}$ m.

Thus, width of the river is $74\sqrt{2}$ meters.

The diagram is given below:

Let $AB$ be the observer, $1.5$ m tall. $CE$ be the tower. Draw line $BD$ parallel to $AC$ which will make $CD = 1.5$ m. In right angle triangle $BDE$ angle of elevation $\angle B = 45^\circ$. Given, $BD = 28.5$ m. Thus,

$\tan45^\circ = \frac{DE}{BD} \Rightarrow DE = 28.5$ m. $\therefore CE = CD + DE = 1.5 + 28.5 = 30$ m.

The diagram is given below:

$BD$ is the pole and $AC$ is the ladder. $C$ is the point which the electrician need to reach to repair the pole which is $1.3$ m below the top of the pole. Total height of the pole is $4$ m, thus, $BC = 4 - 1.3 = 2.7$ m.

We are given than ladder makes an angle of $60^\circ$ with the horizontal.

$\therefore \sin60^\circ = \frac{BC}{AC} \Rightarrow AC = \frac{BC}{\sin60^\circ} = \frac{2.7\sqrt{3}}{2} = 3.12$ m.

The diagram is given below:

$A$ is the point of observation. $B$ is the foot of the tower and $C$ is the top of the tower. $CD$ is the height of the water tank above the tower. Given $AB = 40$ m .

In $\triangle ABC, \tan30^\circ = \frac{BC}{AB} \Rightarrow BC = AB.\tan30^\circ = \frac{40}{\sqrt{3}} = 23.1$ m, which is height of the toweer.

In $\triangle ABD, \tan45\circ = \frac{BD}{AB} \Rightarrow BD = AB.\tan45^\circ = 40.1 = 40$ m which is combined height of the tower and water tank. Thus, height or depth of the water tank $= CD = BD - BC = 40 - 23.1 = 16.9$ m.

The diagram is given below:

The shaded region is the river. $AB$ is the tree and $C$ is the initial point of the observer. $D$ is the final point of observation. Given, $CD = 20$ m. Let $AB = h$ m and $AC = x$ m.

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

In $\triangle ABD \tan30^\circ = \frac{h}{x + 20}\Rightarrow \sqrt{3}h = x + 20$

$\Rightarrow 3x = x + 20 \Rightarrow x = 10$ m. $\Rightarrow h = 10\sqrt{3}$ m.

The diagram is given below:

$AC$ is the tree before breaking. Portion $BC$ has borken and has become $BD$ which makes an angle of $60^\circ$ with remaining portion of tree standing. If $AB = x$ m, then $BD = 12 - x$ because original height of the tree is given as $12$ m.

In $\triangle ABD, \sin60^\circ = \frac{x}{12 - x} \Rightarrow \frac{\sqrt{3}}{2} = \frac{x}{12 - x} \Rightarrow x = 5.57$

The diagram is given below:

$AC$ is the tree before breaking. Portion $BC$ has borken and has become $BD$ which makes an angle of $30^\circ$ with remaining portion of tree standing. If $AB = x$ m, then $BD = l - x$ where $l$ is the original height of the tree.

In $\triangle ABD, \sin30^\circ = \frac{x}{l - x} = \frac{1}{2} \Rightarrow 3x = l$

$\cos30^\circ = \frac{30}{l - x} = \frac{\sqrt{3}}{2} \Rightarrow x = 17.32 \Rightarrow l = 51.96$ m.

The diagram is given below:

$AB$ is the tower. Initial observation point is $D$ where angle of elevation is $\alpha$ such that $\tan\alpha = \frac{5}{12}$. $C$ is the second point of observation where angle of elevation is $\beta$ such that $\tan\beta = \frac{3}{4}$. Given, $CD = 192$ meters. Let $h$ be the height of the tower and $x$ be the distance of $C$ from the foot of the tower i.e. $A$.

In $\triangle ABC, \tan\beta = \frac{3}{4} = \frac{h}{x}$

In $\triangle ABD, \tan\alpha = \frac{5}{12} = \frac{h}{x + 192} \Rightarrow h = 180$ meters.

The diagram is given below:

$AB$ is the tower. When the sun’s altittude is $45^\circ$ the shadow reached $C$. When the shadow reached the altitude of sun becomes $30^\circ$. Let $h$ meters be the height and $x$ meters be the distance of of initial point of observation from foot of the tower. Given $CD = 10$ meters.

In $\triangle ABC, \tan45^\circ = 1 = \frac{h}{x}\Rightarrow x = h$

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{x + 10}\Rightarrow h = \frac{10}{\sqrt{3} - 1} = 13.66$ meters.

The diagram is given below:

This problem is same as previous problem, where $10$ m is replaced by $1$ km. Processing similarly, we obtain $h = 1.366$ km.

The diagram is given below:

This problem is same as two previous problems. The height of the mountain is $5.071$ km.

The diagram is given below:

This problem is same as $11$-th. Proceeding similarly, we find width of river as $20$ m and height of the tree as $20\sqrt{3}$ m.

The diagram is given below:

Height of the plane is $1200$ m which is $AB$. The ships are located at $C$ and $D$. Let $CD = d$ m and $AC = x$ m.

In $\triangle ABC, \tan60^\circ = \frac{1200}{x} \Rightarrow x = \frac{1200}{sqrt{3}} = 400\sqrt{3}$ m.

In $\triangle ABC, \tan30^\circ = \frac{1200}{x + d} \Rightarrow x + d = 1200\sqrt{3} \Rightarrow d = 800\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the flag staff having height $h$ and $AC$ be the shadow when sun’s altitude is $60^\circ$. Let $AD$ be the shadow when sun’s altitude is $\theta^\circ$. If we let $AC = x$ m then $AD = 3x \Rightarrow CD = 2x$.

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

In $\triangle ABD \tan\theta = \frac{h}{3x} = \frac{1}{\sqrt{3}}\Rightarrow \theta = 30^\circ$.

The diagram is given below:

Let $AB$ be the height of the plane, equal to $200$ m. Let the shaded region present the river such that width $CD = x$ m.

In $\triangle ABD, \tan45^\circ = \frac{200}{AD} \Rightarrow AD = 200$ m.

Clearly, $AC = 200 - x$ m. In $\triangle ABC, \tan60^\circ = \frac{200}{200 - x} \Rightarrow x = 84.53$ m.

The diagram is given below:

Let $AC$ and $BD$ represent the towers having height $h$. Given the distance between towers is $100$ m which is $CD$. Let the point of observation be $E$ which is at distance $x$ from $C$ and $100 - x$ from $D$. Angle of elevations are given as $30^\circ$ and $60^\circ$.

In $\triangle ACE, \tan60^\circ = \sqrt{3} = \frac{h}{x}\Rightarrow h = \sqrt{3}x$.

In $\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \Rightarrow x = 25, h = 25\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the light house, $C$ and $D$ are the two locations of the ship. The height of the light house is given as $100$ m. The angle of elevations are given as $30^\circ$ and $45^\circ$. Let $AC = y$ m and $CD = x$ m.

In $\triangle ABC, \tan45^\circ = 1 = \frac{100}{y} \Rightarrow y = 100$.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{100}{x + y} = \frac{100}{100 + x} \Rightarrow x = 73.2$ m.

The diagram is given below:

The diagram represents the top $PQ$ and $XY$ as given in the problem. The angle of elevations are also given. Draw $YZ$ parallel to $ZQ$ and thus, $PZ = 40$ m. Let $ZQ = x$.

In $\triangle QYZ, \tan45^\circ = 1 = \frac{ZQ}{YZ} \Rightarrow YZ = x$ m. Thus, $PX = x$ m.

In $\triangle PQX, \tan60^\circ = \sqrt{3} = \frac{x + 40}{x} \Rightarrow x = \frac{40}{\sqrt{3} - 1}$ m.

Height of toewr is $x + 40 = \frac{40\sqrt{3}}{\sqrt{3} - 1}$

In $\triangle PQX, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{PQ}{XQ} \Rightarrow XQ = \frac{80}{\sqrt{3} - 1}$ m.

The diagram is given below:

Let $AB$ and $CD$ are the houses. Given $CD = 15$ m. Let the width of the street is $AC = ED = x$ m. The angle of depression and elevation are given as $45^\circ$ and $30^\circ$ respectively. Draw $ED\parallel AC$.

In $\triangle ACD, \tan45^\circ = 1 = \frac{CD}{AC} \Rightarrow AC = 15$ m. Thus, $ED$ is also $15$ m because $ED$ is paralle to $AC$.

In $\triangle BED, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{ED} \Rightarrow BE = 5\sqrt{3}$ m.

Thus, total height of the house $= 15 + 5\sqrt{3} = 23.66$ m.

The diagram is given below:

Let $AB$ represent the building and $CD$ the tower. Let $CD = h$ m and given $AB = 60$ m. Also, let $AC = x$ m. Draw $DE\parallel AC$, thus $CE = x$ m and $AE = h$ m.

The angles of depression are given which would be same as angle of elevation from top and bottom of tower.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{60}{x} \Rightarrow x = 20\sqrt{3}$ m.

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

$\therefore$ Height of the building $CD = AE = AB - BE = 60 - 20 = 40$ m.

The diagram is given below:

Let $CD$ represent the deck of the ship with height $10$ m and $AB$ the hill. The water level is $AC$. Draw $DE||AC$ and let $AC = DE = x$ m.

The angle of elevation are shown as given in the question.

In $\triangle ACD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{CD}{x} \Rightarrow x = 10\sqrt{3}$ m.

In $\triangle BDC, \tan60^\circ = \sqrt{3} = \frac{BE}{x} \Rightarrow BE = 30$ m.

Thus, height of the hill $= AE + BE = 10 + 30 = 40$ m.

The diagram is given below:

Let $CE$ be the line in which plane is flying and $ABD$ be the horizontal ground. Since the plane is flying at a constant height of $3600\sqrt{3}$ m, we have $BC = DE = 3600\sqrt{3}$ m. Let $AB = x$ m and $BD = y$ m.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{3600\sqrt{3}}{x} \Rightarrow x = 3600$ m.

In $\triangle ADE, \tan40^\circ = \frac{1}{\sqrt{3}} = \frac{3600\sqrt{3}}{x + y} \Rightarrow y = 7200$ m.

Thus, the plane flies $7200$ m in $30$ s. Speed of plane $= \frac{7200}{30}.\frac{3600}{1000} = 284$ km/hr.

The diagram is given below:

Let $AC$ be the river and $BD$ be the tree on the island in the river. Given wdith of the river $AC$ as $100$ m. Let $BC = x$ m $\Rightarrow AB = 100 - x$ m. The angles of elevation are shown as given in the question. Let $BD = h$ m be the height of the tower.

In $\triangle BCD, \tan45^\circ = 1 = \frac{h}{x} \Rightarrow h = x$ m.

In $\triangle ABC, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{h}{100 - x} \Rightarrow x = \frac{100}{\sqrt{3} + 1} = h$ m.

The diagram is given below:

Let $AB$ be the first tower and $CD$ be the second tower. Given $AC = 140$ m and $CD = 40$ m. Let $AC$ be the horizontal plane. Draw $DE\parallel AC \Rightarrow DE = 140$ m and $AE = 60$ m. Angle of elevation is shown as given in the question from top of second tower to top of first tower to be $30^\circ$.

In $\triangle BDE, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{BE}{140} \Rightarrow BE = \frac{140}{\sqrt{3}}$ m.

Thus, total height of first tower is $\frac{140}{\sqrt{3}} + 60$ m.

The diagram is given below:

Let $AD$ be the horizontal ground. Let $AB$ and $AC$ be the heights at which planes are flying. Given $AC = 4000$ m. Also, given are angles of elevation of the two aeroplanes. Let point of observation be $D$ and $AD = b$ m.

In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{AC}{AD}\Rightarrow b = \frac{4000}{\sqrt{3}}$ m.

In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} \Rightarrow AB = b = \frac{4000}{\sqrt{3}}$ m.

Therefore, distance between heights of two planes $= 4000.\frac{\sqrt{3} - 1}{\sqrt{3}}$ m.

The diagram is given below:

Let $BC$ be the tower where $B$ is the foot of the toewr. Let $A$ be the point of observation. Given $\angle BAC = 60^\circ$.

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

The diagram is given below:

Let $BC$ be the wall and $AC$ the ladder. Given distance of the foot of the ladder is $9.5$ m away from the wall i.e. $AB = 9.5$ m. The angle of elevation is given as $\angle BAC = 60^\circ$.

In $\triangle ABC, \cos60^\circ = \frac{1}{2} = \frac{AB}{AC} \Rightarrow AC = 19$ m.

The diagram is given below:

Let $BC$ be the wall and $AC$ the ladder. Given distance of the foot of the ladder is $2$ m away from the wall i.e. $AB = 2$ m. The angle of elevation is given as $\angle BAC = 60^\circ$.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{BC}{AC} \Rightarrow BC = 2\sqrt{3}$ m.

The diagram is given below:

Let $BC$ be the electric pole, having a height of $10$ m. Let $AC$ be the length of wire. The angle of elevation is given as $\angle BAC=45^\circ$.

In $\triangle ABC, \sin45^\circ = \frac{1}{\sqrt{2}} = \frac{BC}{AC} \Rightarrow AC = 10\sqrt{2}$ m.

The diagram is given below:

Let $BC$ represent the height of kite. Given $BC = 75$ m. Let $AC$ represent the length of the string. The angle of elevation is given as $60^\circ$.

In $\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} \Rightarrow AC = 50\sqrt{3}$ m.

The diagram is given below:

Let $BC$ represent the wall and $AC$ the ladder. Given that the length of ladder is $15$ m. The angle of elevation of the wall from foot of the tower is given as $60^\circ \Rightarrow \angle BAC = 60^\circ$.

In $\triangle ABC, \sin60^\circ = \frac{\sqrt{3}}{2} = \frac{BC}{AC} \Rightarrow BC = \frac{15\sqrt{3}}{2}$ m.

The diagram is given below:

Let $BC$ be the tower and $CD$ be the flag staff, the heights of which are to be found. Let $A$ be the point of obsevation. Given that $AB = 70$ m. The angle of elevation of the foot and the top of flag staff are given as $45^\circ$ and $60^\circ$ i.e. $\angle BAC = 45^\circ$ and $\angle BAD = 60^\circ$.

In $\triangle ABC, \tan45^\circ = 1 = \frac{BC}{AB} \Rightarrow BC = 70$ m, which is height of the tower.

In $\triangle ABD, \tan60^\circ = \sqrt{3} = \frac{BD}{AB} \Rightarrow BD = 70\sqrt{3}$ m, which is combined height of tower and flag staff. Thus, $CD = 70(\sqrt{3} - 1)$ m, which is height of flag staff.

This problem is same as 12. Put $15$ instead of $12$.

The diagram is given below:

Let $AB$ be the tower and $BC$ the flag staff, whose height is $5$ m. Let $D$ be the point of observation. Given that angle of elevation of the foot of the flag staff is $30^\circ$ and that of top is $60^\circ$ i.e. $\angle ADB = 30^\circ$ and $\angle ADC = 60^\circ$. Let $AB = h$ m and $AD = x$ m.

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

In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{h + 5}{x} \Rightarrow h = 2.5~{\rm m}, x = 2.5\sqrt{3}$ m.

This problem is same as 15. Put $50$ m instead of $10$ m and $60^\circ$ instead of $45^\circ$.

This problem is similar to 15. Put $45^\circ$ instead of $30^\circ$ and $60^\circ$ instead of $30^\circ$.

The diagram is given below:

Let $AB$ be the current height of the skydiver as $h$ m. $C$ and $D$ are two points observed at angle of depression $45^\circ$ and $60^\circ$ which woule be equal to angle of elevation from these points. Given that $CD = 100$ m. Let $AC = x$ m.

In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} = \frac{h}{x + 100} \Rightarrow h = x + 100$ m.

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

$\Rightarrow x = \frac{100}{\sqrt{3} - 1}, h = \frac{100\sqrt{3}}{\sqrt{3} - 1}$ m.

The diagram is given below:

Let $AB$ be the tower having a height of $150$ m. Let $C$ and $D$ be the points observed such that $\angle ADB = 45^\circ$ and $\angle ACB = 60^\circ$. Let $AC = y$ m and $CD = x$ m. We have to find $x$.

In $\triangle ABC, \tan60^\circ = \sqrt{3} = \frac{AB}{AC} = \frac{150}{y} \Rightarrow y = 50\sqrt{3}$ m.

In $\triangle ABD, \tan45^\circ = 1 = \frac{AB}{AD} = \frac{150}{x + y}\Rightarrow x = 150 - 50\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the towerr having a height of $h$ m. Let $C$ and $D$ be the points observed such that $\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$. Let $AC = x$ m. Given $CD = 150$ m.

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

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{150}{x + 150} \Rightarrow h = 75\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the towerr having a height of $h$ m. Let $C$ and $D$ be the points observed such that $\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$. Let $AC = x$ m. Given $CD = 100$ m.

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

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AD} = \frac{h}{x + 100} \Rightarrow x = 50$ m.

Thus, $h = 50\sqrt{3}$ m. Distance of initial point $= x + 100 = 150$ m.

The diagram is given below:

Let $AB$ be the tower and $CD$ be the building. Given $CD = 15$ m. $AC$ is the horizontal plane joining foot of the building and foot of the tower having width $x$ m. Draw $DE||AC$ then $DE = x$ m and $AE = 15$ m.

In $\triangle BDE, \tan30^\circ = \frac{h}{x}\Rightarrow x = \sqrt{3}h$ m.

In $\triangle ABC, \tan60^\circ = \frac{h}{x + 15}\Rightarrow h = 7.5$ m and $x = 7.5\sqrt{3}$ m.

The diagram is given below:

Let $AB$ be the tower and $BC$ be the flag staff having heights $x$ and $y$ m respectively. The distance of foot of tower from the point of observation $9$ m. The angles of elevation of the foot and the top of the flag staff are $30^\circ$ and $60^\circ$ as given in the question.

In $\triangle ABD, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{x}{9}\Rightarrow x = 3\sqrt{3}$ m.

In $\triangle ACD, \tan60^\circ = \sqrt{3} = \frac{x + y}{9} \Rightarrow y = 6\sqrt{3}$ m.

The diagram is given below:

Let $AC$ be the full tree and $BC$ is the portion which has fallen. $BC$ becomes $BD$ after falling and angle of elevation is $30^\circ$. Let the height of remaining portion of tree be $AB = x$ m. Also, $BC = BD = 8$ m.

In $\triangle ABC, \sin30^\circ = \frac{1}{2} = \frac{AB}{BD} \Rightarrow AB = 4$ m.

Thus, total height of the tree is $12$ m.

The diagram is given below:

Let $AB$ be the building with height $10$ m. Let $BC$ be the flag with height $h$ m. Also, let distance between $P$ and foot of the building as $AP = x$ m. The angle of elevation of top of the building is $30^\circ$ and that of the flag is $45^\circ$.

In $\triangle ABP, \tan30^\circ = \frac{1}{\sqrt{3}} = \frac{AB}{AP} = \frac{10}{x} \Rightarrow x = 10\sqrt{3}$ m.

In $\triangle ACP, \tan45^\circ = 1 = \frac{AC}{AP} = \frac{10 + h}{x} \Rightarrow h = 10(\sqrt{3} - 1)$ m.