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Neokaal Study

Some Applications of Trigonometry

Practice Worksheet

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

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  1. 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.

  2. 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.

  3. 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.

  4. From a point 40 m from a tower, its angle of elevation is $30^\circ$. Find its height.

  5. A taut kite string makes $60^\circ$ with level ground while the kite is 90 m high. Find the string length.

  6. 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?

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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.

  13. From a 120 m lighthouse, two ships on the same side have depression angles $30^\circ$ and $45^\circ$. Find the distance between them.

  14. 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.

  15. 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?

Neokaal Study

Answer Key

  1. $15$ m

    1. The cable is the hypotenuse.
    2. $h=30\sin30^\circ=15$ m.

  2. $12\sqrt{3}$ m

    1. The stump is $d\tan30^\circ$ and the broken part is $d\sec30^\circ$.
    2. Their sum is $\frac{12}{\sqrt3}+\frac{2(12)}{\sqrt3}=12\sqrt3$ m.

  3. $8$ m and $\frac{8}{\sqrt3}$ m

    1. For each slide, $\sin\theta=\text{height}/\text{length}$.
    2. The lengths are $4/\sin30^\circ=8$ and $4/\sin60^\circ=8/\sqrt3$ m.

  4. $\frac{40}{\sqrt3}$ m

    1. Use tangent with opposite height and adjacent distance.
    2. $h=40\tan30^\circ=40/\sqrt3$ m.

  5. $180\!/\sqrt3$ m

    1. The string is the hypotenuse.
    2. $L=90/\sin60^\circ=180/\sqrt3$ m.

  6. $\frac{60}{\sqrt3}$ 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.

  7. $40(\sqrt3-1)$ m

    1. The horizontal distance equals 40 m because $\tan45^\circ=1$.
    2. Total height is $40\sqrt3$ m, so mast height is $40(\sqrt3-1)$ m.

  8. $\frac{6}{\sqrt3-1}$ m

    1. If pedestal height is $p$, its horizontal distance from the point is also $p$.
    2. Then $(p+6)/p=\sqrt3$, so $p=6/(\sqrt3-1)$ m.

  9. $40$ m

    1. The separation is $120/\tan60^\circ$.
    2. Building height $=(120/\sqrt3)\tan30^\circ=120/3=40$ m.

  10. height $=108\sqrt3/4$ m; distances $=108/4$ m and $324/4$ m

    1. If the nearer distance is $x$, equality of heights gives $x\tan60^\circ=(W-x)\tan30^\circ$.
    2. Thus $x=108/4$ and $h=108\sqrt3/4$ m.

  11. width $=15$ m; height $=15\sqrt3$ m

    1. Let the width be $x$. Equate $x\tan60^\circ$ and $(x+shift)\tan30^\circ$.
    2. This gives $x=30/2=15$ m and $h=15\sqrt3$ m.

  12. $14(1+\sqrt3)$ m

    1. The $45^\circ$ depression makes the horizontal separation 14 m.
    2. The tower rises a further $14\sqrt3$ m above the roof.

  13. $120(\sqrt3-1)$ m

    1. Their distances from the foot are $120\sqrt3$ and $120$ m.
    2. Subtract the nearer distance from the farther one.

  14. $\frac{168}{\sqrt3}$ m

    1. The vertical separation is 84 m.
    2. The horizontal distances are $84/\sqrt3$ and $84\sqrt3$; subtract them.

  15. $2$ seconds

    1. For tower height $h$, the two ground distances are $h\sqrt3$ and $h/\sqrt3$.
    2. The first interval covers twice the remaining distance, so the remaining time is 4/2 seconds.