Question

A ladder 20 ft. long reaches a point 20 ft. below the top of a flag. The angle of elevation of the top of the flag at the foot of the ladder is ${60}^o$. Find the height of the flag.

Answer 1

Given that,

A ladder is $20ft$ long.

It touches a point $20ft$ below a flag.

The angle of elevation of the top of the flag at the foot of the ladder is ${60}^o$.

To find out,

The height of the flag.

Based on the given information, we can draw the figure given above.

Here, $DB$ is the required height of the flag.

Also, $AC=20ft,CD=20ft,\angle b={90}^o,\angle BAD={60}^o$

In $\Delta DBA$,

$\angle B+\angle BAD+\angle ADB={180}^o\left[\text{Interior angle sum property}\right]$

Hence, $\angle ADB={180}^o-{90}^o-{60}^o$

$={30}^o$

Now, in $\Delta DCA$,

$CD=AC=20ft$

We know that, angles opposite to equal sides of a triangle are equal.

Hence, $\angle ADC=\angle CAD={30}^0$

Now, $\angle CAB=\angle DAB-\angle DAC$

Hence, $\angle CAB={60}^o-{30}^o$

$={30}^o$

We know that, $\tan \theta =\frac{\text{Opposite Side}}{\text{Adjacent Side}}$

Hence, in $\Delta BCA$,

$\tan {30}^o=\frac{h}{AB}$

$\Rightarrow AB=h\sqrt{3}[\because \tan {30}^o=\frac{1}{\sqrt{3}}]$

Also, in $\Delta DBA$,

$\tan {60}^o=\frac{h+20}{AB}$

But, $AB=h\sqrt{3}$ and $\tan {60}^o=\sqrt{3}$

Hence, $\sqrt{3}=\frac{h+20}{h\sqrt{3}}$

$\Rightarrow 3h=h+20$

$\Rightarrow 2h=20$

$\therefore h=10ft$

Hence, $DB=h+20$

$=10+20$

$=30ft$

Hence, the height of the flag is $30ft$.