Question

Which of the following statements is true?

$I$. The angle of elevation of the top of a hill at the foot of the tower is ${60}^o$ and the angle of elevation of the top of the tower from the foot of the hill is ${30}^o$. If the tower is $50m$ high, then height of the hill is $150m$.
$II$. An aeroplane flying horizontally $1km$ above the ground is observed at an angle ${60}^o$. After $10\text{seconds}$ its elevation changes to ${30}^o$. Then the speed of the aeroplane is $527.04\text{km/h}$.
$III$. A man in a boat rowing away from light house $100m$ high takes $2\text{minutes}$ to change the angle of elevation of the top of the light house from ${60}^o$ to ${30}^o$. Then the speed of the boat is $40\text{m/minute}$.

• A. $I$
• B. $II$
• C. $III$
• D. $I\&III$

Answer 1

The correct option is A $I$

$I$

Let the height of the hill be ${}^{\prime }{h}^{\prime }m$ $\&$ distance between the foot of hill and tower be ${}^{\prime }{x}^{\prime }m.$

Now, In $△ACB,$

$\Rightarrow \tan 60°=\frac{h}{x}$

$\Rightarrow \sqrt{3}=\frac{h}{x}$

$\Rightarrow x=\frac{h}{\sqrt{3}}⟶\left(2\right)$

In $△DBC,$

$\Rightarrow \tan 30°=\frac{50}{x}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{x}$

$\Rightarrow x=50\sqrt{3}$

From $\left(1\right)\xi \left(2\right),$ we get

$\Rightarrow \frac{h}{\sqrt{3}}=50\sqrt{3}$

$\Rightarrow h=150m$

$II$

Let the distance travelled in $2$ min be ${}^{\prime }{y}^{\prime }m.$

${}^{\prime }{x}^{\prime }$ be the distance initially.

Now, In $△EDC,$

$\Rightarrow \tan 60°=\frac{EC}{DC}=\frac{1000}{x}$

$\Rightarrow \sqrt{3}=\frac{1000}{x}$

$\Rightarrow x=\frac{1000}{\sqrt{3}}m⟶\left(1\right)$

Now, In $△ADB,$

$\Rightarrow \tan 30°=\frac{AB}{DB}=\frac{1000}{x+y}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{1000}{\frac{1000}{\sqrt{3}}+y}$ $\left[From\right(1\left)\right]$

Or $1000\sqrt{3}=\frac{1000}{\sqrt{3}}+y$

$\Rightarrow y=1000\sqrt{3}-\frac{1000}{\sqrt{3}}$

Or $y=\frac{3000-1000}{\sqrt{3}}=\frac{2000}{\sqrt{3}}=\frac{2000\sqrt{3}}{3}m$

Now, $y$ is the distance travelled in $2min.$

$\therefore$ Speed of plane $=\frac{Distance}{Time}=\frac{2000\sqrt{3}}{6}m/min.$

$=\frac{2000\sqrt{3}\times 60}{6\times 1000}km/h$

$=20\sqrt{3}km/h$

$III$

Let ${}^{\prime }{y}^{\prime }$ be the initial distance of boat from light house.

${}^{\prime }{x}^{\prime }$ be the distance travelled in $2min.$

Now, In $△ACB,$

$\Rightarrow \tan 60°=\frac{AB}{BC}=\frac{100}{y}$

$\Rightarrow \sqrt{3}=\frac{1000}{y}$

$\Rightarrow y=\frac{100}{\sqrt{3}}m⟶\left(1\right)$

Now, In $△ADB,$

$\Rightarrow \tan 30°=\frac{AB}{BD}=\frac{100}{x+y}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{x+y}$

$\Rightarrow x+y=100\sqrt{3}=x+\frac{100}{\sqrt{3}}=100\sqrt{3}$ $\left[From\right(1\left)\right]$

$\Rightarrow x=100\sqrt{3}-\frac{100}{\sqrt{3}}=\frac{300-100}{\sqrt{3}}=\frac{200}{\sqrt{3}}$

$\Rightarrow x=\frac{200\sqrt{3}}{3}m$

Now, Speed of the boat $=\frac{Distance{}^{\prime }x{}^{\prime }}{Time}=\frac{200\sqrt{3}}{2\times 3}=\frac{100\sqrt{3}}{3}m/min.$

Hence the answer is only $I$ statement is true.