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Neokaal Study

Some Applications of Trigonometry

Competency Practice

Cloud elevation and lake reflection depth, multi-position elevations, and angles of depression.

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  1. The angle of elevation of a cloud from a point $20\text{ m}$ above the surface of a lake is $30^\circ$, and the angle of depression of the reflection of the cloud in the lake from the same point is $60^\circ$. Find the height of the cloud above the surface of the lake.

  2. An observer's eye level is 2 m. Looking at a 32 m building, the angle of elevation changes from $30^\circ$ to $60^\circ$ after walking straight towards it. How far did the observer walk?

  3. A transmission mast stands on a 50 m building. From a ground point, the building roof and mast top have elevation angles $45^\circ$ and $60^\circ$. Find the mast's height.

  4. From a 105 m lighthouse, two ships on the same side have depression angles $30^\circ$ and $45^\circ$. Find the distance between them.

  5. A tree breaks and its top touches the ground 18 m from the foot, making $30^\circ$ with the ground. Find the tree's original height.

  6. A taut kite string makes $60^\circ$ with level ground while the kite is 72 m high. Find the string length.

Neokaal Study

Answer Key

  1. $40\text{ m}$

    1. Let $H$ be the height of the cloud above the lake surface. The observation point $P$ is at height $h = 20\text{ m}$ above the lake.
    2. The vertical height of the cloud above $P$ is $(H - 20)\text{ m}$, and the depth of the reflection below $P$ is $(H + 20)\text{ m}$.
    3. From the angle of elevation ($30^\circ$): $\tan 30^\circ = \frac{H - 20}{x} \implies x = (H - 20)\sqrt{3}$.
    4. From the angle of depression ($60^\circ$): $\tan 60^\circ = \frac{H + 20}{x} \implies x = \frac{H + 20}{\sqrt{3}}$.
    5. Equating expressions for $x$: $3(H - 20) = H + 20 \implies 2H = 80 \implies H = 40\text{ m}$.

  2. $\frac{60}{\sqrt3}$ m

    1. The vertical rise above eye level is 30 m.
    2. The two distances are $30\sqrt3$ and $30/\sqrt3$; their difference is $60/\sqrt3$ m.

  3. $50(\sqrt3-1)$ m

    1. The horizontal distance equals 50 m because $\tan45^\circ=1$.
    2. Total height is $50\sqrt3$ m, so mast height is $50(\sqrt3-1)$ m.

  4. $105(\sqrt3-1)$ m

    1. Their distances from the foot are $105\sqrt3$ and $105$ m.
    2. Subtract the nearer distance from the farther one.

  5. $18\sqrt{3}$ m

    1. The stump is $d\tan30^\circ$ and the broken part is $d\sec30^\circ$.
    2. Their sum is $\frac{18}{\sqrt3}+\frac{2(18)}{\sqrt3}=18\sqrt3$ m.

  6. $144\!/\sqrt3$ m

    1. The string is the hypotenuse.
    2. $L=72/\sin60^\circ=144/\sqrt3$ m.