Question

A T.V tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60{}^\circ $. From a point 20 m away from this point on the same bank, the angle of elevation of the top of the tower is $30{}^\circ $. Find the height of the tower and the width of the river.

Answer 1

Hint: Assume the height of the tower be $h$. Use the fact that $\tan \alpha=\dfrac{A D}{A B}$ to find the length $\mathrm{AB}$ in terms of $\mathrm{h}$. Use the fact that $\tan \beta=\dfrac{A D}{A C}$ and hence form an equation in $\mathrm{h}$. Solve for $\mathrm{h}$. The value of $\mathrm{h}$ gives the height

of the tower $(A D)$. Hence find the width of the river $(A B)$.

Complete step-by-step answer:

Given: $A D$ is a tower on the bank of a river. $B$ is another point directly opposite to the foot of the tower on the

other bank of the river. The angle of elevation of the top of the tower from point $\mathrm{B}$ is $\alpha=60^{\circ}$. From point $\mathrm{C}$,

the angle of elevation of the top of the tower is $\beta=30^{\circ} . \mathrm{BC}=20 \mathrm{~m}$

In triangle $A B D, A D$ is the side opposite to $\alpha$ and $A B$ is the side adjacent to $\alpha$.

We know that in a triangle $\tan \theta=\dfrac{\text { Opposite side }}{\text { Adjacent side }}$. Hence, we have $\tan \alpha=\dfrac{A D}{A B}$

Multiplying both sides of the equation by $\dfrac{A B}{\tan \alpha}$, we get $A B=\dfrac{A D}{\tan \alpha}=\dfrac{h}{\tan \alpha}(i)$

Also, in triangle $A C D$, we have $A D$ is the side opposite to $\beta$ and $A C$ is the side adjacent to $\beta$

We know that in a triangle $\tan \theta=\dfrac{\text { Opposite side }}{\text { Adjacent side }}$. Hence, we have $\tan \beta=\dfrac{A D}{A C}$

Multiplying both sides by $\dfrac{A C}{\tan \beta}$, we get

$A C=\dfrac{AD}{\tan \beta}=\dfrac{h}{\tan \beta}(i i)$

Subtracting equation (ii) from equation (i), we get $A C-A B=\dfrac{h}{\tan \beta}-\dfrac{h}{\tan \alpha}$

Since $A C-A B=B C$

Hence, we have $B C=h\left(\dfrac{1}{\tan \beta}-\dfrac{1}{\tan \alpha}\right)$

Put $\mathrm{BC}=20 \mathrm{~m}, \beta=30^{\circ}$ and $\alpha=60^{\circ}$, we get

$20=h\left(\dfrac{1}{\tan 30^{\circ}}-\dfrac{1}{\tan 60^{\circ}}\right)$

We know that $\tan 30^{\circ}=\dfrac{1}{\sqrt{3}}$ and $\tan 60^{\circ}=\sqrt{3}$

Hence, we have $20=h\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right)=\dfrac{2 h}{\sqrt{3}}$

Multiplying both sides by $\dfrac{\sqrt{3}}{2}$, we get $20 \times \dfrac{\sqrt{3}}{2}=h \Rightarrow h=10 \sqrt{3}$

Substituting the value of $h$ in equation (i), we get $A B=\dfrac{10 \sqrt{3}}{\tan 60^{\circ}}=10$

Hence the height of the tower is $10 \sqrt{3} m$, and the width of the river is $10 \mathrm{~m}$.

Note: Verification:

We have $A D=10 \sqrt{3}$ and $A B=10$

Hence, we have $\dfrac{A D}{A B}=\dfrac{10 \sqrt{3}}{10}=\sqrt{3}=\tan 60^{\circ}$

Hence, we have $\alpha=60^{\circ}$.

Also, we have $A C=A B+B C=20+10=30 m$ Hence, we have $\dfrac{A D}{A C}=\dfrac{10 \sqrt{3}}{30}=\dfrac{1}{\sqrt{3}}=\tan 30^{\circ}$

Hence, we have $\beta=30^{\circ}$.

Hence our answer is verified to be correct.