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Some Applications of Trigonometry: Competency Practice

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

6 problems·20–25 min·★★★★★
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  1. Problem 1
    applications-of-trigonometry·★★★★★
    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.
    ▶Answer
    $40\text{ m}$
    ▶Step-by-step solution
    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 $x$: $3(H - 20) = H + 20 \implies 2H = 80 \implies H = 40\text{ m}$.
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  2. Problem 2
    heights-and-distances·★★★★☆
    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?
    ▶Answer
    $\frac{60}{\sqrt3}$ m
    ▶Step-by-step solution
    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.
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  3. Problem 3
    heights-and-distances·★★★★☆
    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.
    ▶Answer
    $50(\sqrt3-1)$ m
    ▶Step-by-step solution
    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.
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  4. Problem 4
    heights-and-distances·★★★★☆
    From a 105 m lighthouse, two ships on the same side have depression angles $30^\circ$ and $45^\circ$. Find the distance between them.
    ▶Answer
    $105(\sqrt3-1)$ m
    ▶Step-by-step solution
    1. Their distances from the foot are $105\sqrt3$ and $105$ m.
    2. Subtract the nearer distance from the farther one.
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  5. Problem 5
    heights-and-distances·★★★☆☆
    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.
    ▶Answer
    $18\sqrt{3}$ m
    ▶Step-by-step solution
    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.
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  6. Problem 6
    heights-and-distances·★★★☆☆
    A taut kite string makes $60^\circ$ with level ground while the kite is 72 m high. Find the string length.
    ▶Answer
    $144\!/\sqrt3$ m
    ▶Step-by-step solution
    1. The string is the hypotenuse.
    2. $L=72/\sin60^\circ=144/\sqrt3$ m.
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