Question

the angle of elevation from point a on the ground to the top of a water tower is 30°. from point b, which is 10 m closer to the tower than point a, the angle of elevation is 45°. determine the height of the water tower. include a diagram in your answer.

Explanation:

Step1: Set up variables

Let the height of the water - tower be $h$ meters and the distance from point $B$ to the base of the tower be $x$ meters.

Step2: Use tangent function for point B

Since the angle of elevation from point $B$ is $45^{\circ}$, and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, we have $\tan45^{\circ}=\frac{h}{x}$. Since $\tan45^{\circ} = 1$, then $h = x$.

Step3: Use tangent function for point A

The distance from point $A$ to the base of the tower is $(x + 10)$ meters. The angle of elevation from point $A$ is $30^{\circ}$. So $\tan30^{\circ}=\frac{h}{x + 10}$. Since $\tan30^{\circ}=\frac{1}{\sqrt{3}}$, we get $\frac{1}{\sqrt{3}}=\frac{h}{h + 10}$ (because $x = h$).

Step4: Solve the equation

Cross - multiply the equation $\frac{1}{\sqrt{3}}=\frac{h}{h + 10}$:
\[h + 10=h\sqrt{3}\]
\[10=h\sqrt{3}-h\]
\[10=h(\sqrt{3}-1)\]
\[h=\frac{10}{\sqrt{3}-1}\]
Rationalize the denominator: $h=\frac{10(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{10(\sqrt{3}+1)}{3 - 1}=\frac{10(\sqrt{3}+1)}{2}=5(\sqrt{3}+1)\approx5(1.732 + 1)=5\times2.732 = 13.66$ meters.

Answer:

$5(1+\sqrt{3})\text{ m}\approx13.66\text{ m}$