2.0 Questions

Question 1. Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are ${30^ \circ }$ and ${45^ \circ }$ respectively. If the lighthouse is $100m$ high. What is the distance between the two ships?

Solution: Let $AB$ be the lighthouse and $C$ and $D$ be the positions of the ship.

so we need to calculate $CD$

$AB = 100m$, $\angle ACB = {30^ \circ }$, $\angle ADB = {45^ \circ }$

$$\begin{equation} \begin{aligned} AB = 100m \\ \frac{{AB}}{{AC}} = \tan {30^ \circ } = \frac{1}{{\sqrt 3 }} \\ AC = 100\sqrt 3 \\ \frac{{AB}}{{AD}} = \tan {45^ \circ } = 1 \\ \Rightarrow AB = AD = 100 \\\end{aligned} \end{equation} $$

Now, $$\begin{equation} \begin{aligned} CD = AC + AD \\ = 100\sqrt 3 + 100 \\ = 100(\sqrt 3 + 1) \\ = 100 \times 2.73 \\ = 273 \\\end{aligned} \end{equation} $$

So the distance between two ships is $273m$

Question 2. The angle of elevation of an aeroplane from a point A on the ground is ${60^ \circ }$ . After a flight of $15$ seconds horizontally, the angle of elevation changes to ${30^ \circ }$ . If the aeroplane is flying at a speed of $200m/s$, then find the constant height at which the aeroplane is flying.

Solution : Let $A$ be the point of observation. Let $E$ and $D$ be positions of the aeroplane initially and after $15$ seconds respectively. Let $BE$ and $CD$ denote the constant height at which the aeroplane is flying.

Given that $\angle DAC = {30^ \circ }$ and $\angle EAB = {60^ \circ }$ .

Let $\begin{equation} \begin{aligned} BE = CD = h \\ AB = x \\\end{aligned} \end{equation} $

Thus, distance covered in $15$ seconds, $$\begin{equation} \begin{aligned} ED = 200 \times 15 \\ = 3000m \\\end{aligned} \end{equation} $$

Thus $BC=3000m$

In right angled triangle $ADC$, $$\begin{equation} \begin{aligned} \tan {30^ \circ } = \frac{{CD}}{{AC}} \\ \Rightarrow CD = AC\tan {30^ \circ } \\\end{aligned} \end{equation} $$

Thus, $$h = (x + 3000) \times \frac{1}{{\sqrt 3 }}...(i)$$

In right angled triangle $AEB$,

$$\begin{equation} \begin{aligned} \tan {60^ \circ } = \frac{{BE}}{{AB}} \\ BE = AB\tan {60^ \circ } \\\end{aligned} \end{equation} $$

Thus, $$h = \sqrt 3 x...(ii)$$

From $(i)$ and $(ii)$

$$\begin{equation} \begin{aligned} \sqrt 3 x = (x + 3000) \times \frac{1}{{\sqrt 3 }} \\ 3x = x + 3000 \\ 2x = 3000 \\ x = 1500 \\\end{aligned} \end{equation} $$

And from $(ii)$, we get

$$h = 1500\sqrt 3 $$

So, the constant height at which airplane is flying is $h = 1500\sqrt 3 $

Question 3. From a point $P$ on a level ground, the angle of elevation of the top of tower is ${30^ \circ }$.The tower is $200m$ high. Find the distance of point $P$ from the footof the tower.

Solution: $$\begin{equation} \begin{aligned} \tan {30^ \circ } = \frac{{RQ}}{{PQ}} \\ \frac{1}{{\sqrt 3 }} = \frac{{200}}{{PQ}} \\ PQ = 200\sqrt 3 \\ = 200 \times 1.73 \\ = 346m \\\end{aligned} \end{equation} $$

So the distance of the point $P$ from the foot of the tower is $346m$.

Question 4. Find the angle of elevation of the sun, if the length of shadow of tree is equal to the height of the tree.

Solution: Consider the diagram shown below where $QR$ represents the tree and $PQ$ represents the shadow of the tree.

We have $$\begin{equation} \begin{aligned} QR = PQ \\ \angle QPR = \theta \\ \tan \theta = \frac{{QR}}{{PQ}} \\\end{aligned} \end{equation} $$

Since $QR=PQ$

So $$\theta = {45^ \circ }$$

Therefore, the angle of elevation is ${45^ \circ }$

Question 5. An observer $2m$ tall is $10\sqrt 3 $ away from a tower. The angle of elevation from his eye to the top of the tower is ${30^ \circ }$. Find the height of the tower.

Solution:$$\begin{equation} \begin{aligned} SR = PQ = 2m \\ PS = QR = 10\sqrt 3 \\ \tan {30^ \circ } = \frac{{TS}}{{PS}} \\ \frac{1}{{\sqrt 3 }} = \frac{{TS}}{{10\sqrt 3 }} \\ TS = 10m \\ TR = TS + SR = 10 + 2 = 12m \\\end{aligned} \end{equation} $$

Height of the tower is $12m$.

Question 6. A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes $8$ minutes for the angle of depression to change from ${30^ \circ }$ to ${45^ \circ }$, how soon after this will the car reach the observation tower.

Solution: Consider the figure $6$ as shown. Let $AB$ be the tower and $C$ and $D$ are the positions of the car.

Then, $$\angle ADC = {30^ \circ },\angle ACB = {45^ \circ }$$

Let $$\begin{equation} \begin{aligned} AB = h,BC = x,CD = y \\ \tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x} \\ \Rightarrow 1 = \frac{h}{x} \\ h = x...(i) \\ \tan {30^ \circ } = \frac{{AB}}{{BD}} = \frac{{AB}}{{(BC + CD)}} = \frac{h}{{x + y}} \\ \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}} \\ x + y = \sqrt 3 h \\ y = \sqrt 3 h - x \\ y = \sqrt 3 h - h \\\end{aligned} \end{equation} $$ (By substituting the value of $x$ by $h$ from $(i)$)

$$y = (\sqrt 3 - 1)h$$

Given that distance $y$ is covered in $8$ minutes, i.e distance $y = (\sqrt 3 - 1)h$ is covered in $8$ minutes

Time to travel $x$ = time to travel $h$ because $x=h$.

Let distance $h$ is covered in $t$ minutes. Since the distance is proportional to the time when speed is constant we have

$$\begin{equation} \begin{aligned} (\sqrt 3 - 1)h \propto 8...(A) \\ h \propto t...(B) \\\end{aligned} \end{equation} $$

Dividing $A$ by $B$ we get,

$$\begin{equation} \begin{aligned} \frac{{(A)}}{{(B)}} = \frac{{(\sqrt 3 - 1)h}}{h} = \frac{8}{t} \\ (\sqrt 3 - 1) = \frac{8}{t} \\ t = \frac{8}{{(\sqrt 3 - 1)}} \\ t = \frac{8}{{1.73 - 1}} \\ t = \frac{8}{{.73}} = \frac{{800}}{{73}} \\\end{aligned} \end{equation} $$

So the time required to reach the observation tower is approximately is $10$ minutes $57$seconds.

Question 7. The top of a $15m$ high tower makes an angle of elevation of ${60^ \circ }$ with the bottom of an electic pole and the angle of elevation with the top of the pole. What is the height of the electric pole?

Solution: Consider the figure shown below where $AC$ represents the tower and $DE$ represents the pole.

Given that $$AC = 15m,\angle ACB = {30^ \circ },\angle AEC = {60^ \circ }$$

Let $$\begin{equation} \begin{aligned} DE = h \\ BC = DE = h \\ AB = 15 - h \\ BD = CE \\ \tan {60^ \circ } = \frac{{AC}}{{CE}} \\ \sqrt 3 = \frac{{15}}{{CE}} \\ CE = \frac{{15}}{{\sqrt 3 }}...(i) \\\end{aligned} \end{equation} $$

$$\begin{equation} \begin{aligned} \tan {30^ \circ } = \frac{{AB}}{{BD}} \\ \frac{1}{{\sqrt 3 }} = \frac{{15 - h}}{{BD}} \\ \frac{1}{{\sqrt 3 }} = \frac{{15 - h}}{{\left( {\frac{{15}}{{\sqrt 3 }}} \right)}} \\\end{aligned} \end{equation} $$ Since $BD=CE$ so by substituting the value of $CE$ from equation $(i)$

$$\begin{equation} \begin{aligned} 15 - h = \left( {\frac{{15}}{{\sqrt 3 }}} \right) \times \frac{1}{{\sqrt 3 }} = \frac{{15}}{3} = 5 \\ h = 15 - 5 \\ h = 10m \\\end{aligned} \end{equation} $$

So the height of the electric pole is $10m$.