Question

1. the cn tower is 553.33 m high. ellen looks up at the top of the tower at a 15° angle of elevation. what is the distance between ellen and the base of the tower to the nearest metre?

Explanation:

Step1: Set up the tangent - ratio equation

We know that in a right - triangle (where the height of the tower is the opposite side and the distance from Ellen to the base of the tower is the adjacent side with respect to the angle of elevation), $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Let the height of the tower $h = 553.33$ m be the opposite side and the distance from Ellen to the base of the tower be $d$ (adjacent side), and $\theta = 15^{\circ}$. So, $\tan(15^{\circ})=\frac{553.33}{d}$.

Step2: Solve for $d$

We can re - arrange the equation to get $d=\frac{553.33}{\tan(15^{\circ})}$. We know that $\tan(15^{\circ})=\tan(45^{\circ}- 30^{\circ})=\frac{\tan45^{\circ}-\tan30^{\circ}}{1 + \tan45^{\circ}\tan30^{\circ}}=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=2-\sqrt{3}\approx0.2679$. Then $d=\frac{553.33}{0.2679}\approx2066$ m.

Answer:

$2066$ m