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 water - tower be $x$ meters.

Step2: Use tangent function for point B

Since the angle of elevation from point $B$ is $45^{\circ}$, we know that $\tan45^{\circ}=\frac{h}{x}$. Since $\tan45^{\circ}=1$, we have $h = x$.

Step3: Use tangent function for point A

The distance from point $A$ to the base of the water - 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: Cross - multiply and solve for h

Cross - multiplying the equation $\frac{1}{\sqrt{3}}=\frac{h}{h + 10}$ gives $h + 10=h\sqrt{3}$.
Rearranging the terms: $h\sqrt{3}-h = 10$.
Factor out $h$: $h(\sqrt{3}-1)=10$.
Then $h=\frac{10}{\sqrt{3}-1}$.
Rationalize the denominator: $h=\frac{10(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$.
Since $(\sqrt{3}-1)(\sqrt{3}+1)=3 - 1=2$, we have $h = 5(\sqrt{3}+1)\approx5(1.732 + 1)=5\times2.732 = 13.66$ meters.

Answer:

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