Trigonometry - Heights and Distances

The angle of elevation of the top of a tower from a point on the ground, which is $30$ m away from the foot of the tower, is $30^{\circ}$. Find the height of the tower.

Let $h$ be the height of the tower $AB$ and $d$ be the distance from the point $C$ to the foot of the tower $B$.

Given: Distance $BC = 30$ m Angle of elevation $\angle ACB = 30^{\circ}$

In right-angled $\triangle ABC$: $\tan 30^{\circ} = \frac{AB}{BC}$ $\frac{1}{\sqrt{3}} = \frac{h}{30}$ $h = \frac{30}{\sqrt{3}}$ $h = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$ $h = \frac{30\sqrt{3}}{3} = 10\sqrt{3}$ m $h = 10 \times 1.732 = 17.32$ m

We use the tangent ratio because we are given the base (distance from tower) and need to find the perpendicular (height of the tower).

From the top of a $75$ m high lighthouse, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Let $AB$ be the lighthouse of height $75$ m. Let $C$ and $D$ be the positions of the two ships.

In $\triangle ABC$ (closer ship with $45^{\circ}$ angle): $\tan 45^{\circ} = \frac{AB}{BC}$ $1 = \frac{75}{BC} \Rightarrow BC = 75$ m

In $\triangle ABD$ (farther ship with $30^{\circ}$ angle): $\tan 30^{\circ} = \frac{AB}{BD}$ $\frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3}$ m

Distance between ships $CD = BD - BC$: $CD = 75\sqrt{3} - 75$ $CD = 75(\sqrt{3} - 1)$ $CD = 75(1.732 - 1) = 75(0.732) = 54.9$ m

The distance between the ships is the difference between their respective horizontal distances from the foot of the lighthouse. We solve for both distances using the tangent ratio and subtract them.

Let $AB$ be the cliff of height $100$ m. Let $C$ be the initial position of the boat and $D$ be the final position. In $\triangle ABC$: $\tan 45^{\circ} = \frac{AB}{BC} \implies 1 = \frac{100}{BC} \implies BC = 100$ m. In $\triangle ABD$: $\tan 30^{\circ} = \frac{AB}{BD} \implies \frac{1}{\sqrt{3}} = \frac{100}{BD} \implies BD = 100\sqrt{3}$ m. Distance travelled $CD = BD - BC = 100\sqrt{3} - 100 = 100(\sqrt{3} - 1)$ m. $CD = 100(1.732 - 1) = 100(0.732) = 73.2$ m.

We use the properties of right triangles. The first angle gives the initial distance, and the second angle gives the total distance from the cliff. Subtracting the two gives the distance the boat moved.

Let $AB = 20$ m be the pole and $CD$ be the tower. Let $AE$ be the horizontal line from the top of the pole to the tower. In $\triangle ABC$ (where $C$ is the foot of the tower): $\tan 30^{\circ} = \frac{AB}{BC} \implies \frac{1}{\sqrt{3}} = \frac{20}{BC} \implies BC = 20\sqrt{3}$ m. Since $AE = BC = 20\sqrt{3}$ m. In $\triangle ADE$ (where $D$ is the top of the tower): $\tan 60^{\circ} = \frac{DE}{AE} \implies \sqrt{3} = \frac{DE}{20\sqrt{3}} \implies DE = 20\sqrt{3} \times \sqrt{3} = 60$ m. Total height of tower $CD = DE + EC = DE + AB = 60 + 20 = 80$ m.

First, calculate the horizontal distance using the angle of depression to the foot. Then, use that distance and the angle of elevation to find the height of the tower section above the pole's height.