Some Applications of Trigonometry: Practice Worksheet

Heights and distances using angles of elevation and depression in varied configurations.

15 problems·20–25 min·★★★★☆

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Problem 1
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A 30 m support cable runs from the top of a vertical mast to the ground and makes $30^\circ$ with the ground. Find the mast's height.
▶Answer
$15$ m
▶Step-by-step solution
The cable is the hypotenuse.
$h=30\sin30^\circ=15$ m.
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Problem 2
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A tree breaks and its top touches the ground 12 m from the foot, making $30^\circ$ with the ground. Find the tree's original height.
▶Answer
$12\sqrt{3}$ m
▶Step-by-step solution
The stump is $d\tan30^\circ$ and the broken part is $d\sec30^\circ$.
Their sum is $\frac{12}{\sqrt3}+\frac{2(12)}{\sqrt3}=12\sqrt3$ m.
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Problem 3
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Two straight slides rise 4 m and 4 m and make angles $30^\circ$ and $60^\circ$ with the ground, respectively. Find both slide lengths.
▶Answer
$8$ m and $\frac{8}{\sqrt3}$ m
▶Step-by-step solution
For each slide, $\sin\theta=\text{height}/\text{length}$.
The lengths are $4/\sin30^\circ=8$ and $4/\sin60^\circ=8/\sqrt3$ m.
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Problem 4
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From a point 40 m from a tower, its angle of elevation is $30^\circ$. Find its height.
▶Answer
$\frac{40}{\sqrt3}$ m
▶Step-by-step solution
Use tangent with opposite height and adjacent distance.
$h=40\tan30^\circ=40/\sqrt3$ m.
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Problem 5
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A taut kite string makes $60^\circ$ with level ground while the kite is 90 m high. Find the string length.
▶Answer
$180\!/\sqrt3$ m
▶Step-by-step solution
The string is the hypotenuse.
$L=90/\sin60^\circ=180/\sqrt3$ m.
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Problem 6
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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
The vertical rise above eye level is 30 m.
The two distances are $30\sqrt3$ and $30/\sqrt3$; their difference is $60/\sqrt3$ m.
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Problem 7
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A transmission mast stands on a 40 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
$40(\sqrt3-1)$ m
▶Step-by-step solution
The horizontal distance equals 40 m because $\tan45^\circ=1$.
Total height is $40\sqrt3$ m, so mast height is $40(\sqrt3-1)$ m.
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Problem 8
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A 6 m statue stands on a pedestal. From one ground point, the elevation angles of the pedestal top and statue top are $45^\circ$ and $60^\circ$. Find the pedestal height.
▶Answer
$\frac{6}{\sqrt3-1}$ m
▶Step-by-step solution
If pedestal height is $p$, its horizontal distance from the point is also $p$.
Then $(p+6)/p=\sqrt3$, so $p=6/(\sqrt3-1)$ m.
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Problem 9
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From the foot of a 120 m tower, a building top has elevation $30^\circ$. From the building foot, the tower top has elevation $60^\circ$. Find the building height.
▶Answer
$40$ m
▶Step-by-step solution
The separation is $120/\tan60^\circ$.
Building height $=(120/\sqrt3)\tan30^\circ=120/3=40$ m.
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Problem 10
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Equal poles stand at opposite edges of a 108 m road. From a point between them their elevation angles are $60^\circ$ and $30^\circ$. Find each pole's height and the two distances.
▶Answer
height $=108\sqrt3/4$ m; distances $=108/4$ m and $324/4$ m
▶Step-by-step solution
If the nearer distance is $x$, equality of heights gives $x\tan60^\circ=(W-x)\tan30^\circ$.
Thus $x=108/4$ and $h=108\sqrt3/4$ m.
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Problem 11
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From the bank opposite a vertical tower, its elevation is $60^\circ$. From a point 30 m farther away on the same line, it is $30^\circ$. Find the canal width and tower height.
▶Answer
width $=15$ m; height $=15\sqrt3$ m
▶Step-by-step solution
Let the width be $x$. Equate $x\tan60^\circ$ and $(x+shift)\tan30^\circ$.
This gives $x=30/2=15$ m and $h=15\sqrt3$ m.
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Problem 12
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From the roof of a 14 m building, a tower's foot has depression $45^\circ$ and its top has elevation $60^\circ$. Find the tower height.
▶Answer
$14(1+\sqrt3)$ m
▶Step-by-step solution
The $45^\circ$ depression makes the horizontal separation 14 m.
The tower rises a further $14\sqrt3$ m above the roof.
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Problem 13
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From a 120 m lighthouse, two ships on the same side have depression angles $30^\circ$ and $45^\circ$. Find the distance between them.
▶Answer
$120(\sqrt3-1)$ m
▶Step-by-step solution
Their distances from the foot are $120\sqrt3$ and $120$ m.
Subtract the nearer distance from the farther one.
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Problem 14
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A balloon moves horizontally at 86 m. From an observer whose eyes are 2 m high, its elevation changes from $60^\circ$ to $30^\circ$. Find the distance travelled.
▶Answer
$\frac{168}{\sqrt3}$ m
▶Step-by-step solution
The vertical separation is 84 m.
The horizontal distances are $84/\sqrt3$ and $84\sqrt3$; subtract them.
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Problem 15
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From a tower, an approaching vehicle has depression $30^\circ$. After 4 seconds at constant speed its depression is $60^\circ$. How many more seconds will it take to reach the tower?
▶Answer
$2$ seconds
▶Step-by-step solution
For tower height $h$, the two ground distances are $h\sqrt3$ and $h/\sqrt3$.
The first interval covers twice the remaining distance, so the remaining time is 4/2 seconds.
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