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Difficulty: MediumPythagorean Theorem and Special Right Triangles

A vertical flagpole stands on flat, horizontal ground. An observer at point SS on the ground measures the angle of elevation to the top of the flagpole to be 60∘60^\circ. A second observer at point PP on the ground, located 3030 feet further from the base of the flagpole along the same line extending from the base through SS, measures the angle of elevation to the top of the flagpole to be 30∘30^\circ. What is the height of the flagpole, in feet?

  1. 15315\sqrt{3}Answer
  2. B
    30330\sqrt{3}
  3. C
    4545
  4. D
    3030
  5. E
    1515

Answer

The height of the flagpole is 15315\sqrt{3} feet.
Let BB be the base and TT be the top of the flagpole. The right triangle TBSTBS has angles 30∘30^\circ, 60∘60^\circ, and 90∘90^\circ at BB. If BS=xBS = x, then the height TB=x3TB = x\sqrt{3}. In right triangle TBPTBP, the angle at PP is 30∘30^\circ, so the base BP=TB3=(x3)3=3xBP = TB\sqrt{3} = (x\sqrt{3})\sqrt{3} = 3x. Since BP=BS+30BP = BS + 30, we set up 3x=x+303x = x + 30, yielding x=15x = 15. Therefore, the height TB=153TB = 15\sqrt{3} feet.

Step-by-Step Solution

1
Model the scenario using right triangles.
Let BB be the base of the flagpole and TT be the top. Triangle TBSTBS is a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle at BB, and triangle TBPTBP is also a 30∘−60∘−90∘30^\circ-60^\circ-90^\circ right triangle at BB.
The flagpole is perpendicular to the horizontal ground.
2
Express side lengths in terms of base distance x=BSx = BS.
In 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle TBSTBS, TB=x3TB = x\sqrt{3} and ST=2xST = 2x.
The side opposite the 60∘60^\circ angle is 3\sqrt{3} times the side opposite the 30∘30^\circ angle.
3
Set up the equation for triangle TBPTBP.
In 30∘−60∘−90∘30^\circ-60^\circ-90^\circ triangle TBPTBP, the side opposite the 30∘30^\circ angle is TB=x3TB = x\sqrt{3}, so the adjacent side BP=(x3)3=3xBP = (x\sqrt{3})\sqrt{3} = 3x.
The side opposite the 60∘60^\circ angle (BPBP) is 3\sqrt{3} times the side opposite the 30∘30^\circ angle (TBTB).
4
Solve for xx using the given distance SP=30SP = 30.
Since BP=BS+SPBP = BS + SP, we have 3x=x+303x = x + 30, which simplifies to 2x=302x = 30, so x=15x = 15.
The total horizontal distance BPBP is the sum of segments BSBS and SPSP.
5
Calculate the height TBTB.
TB=153TB = 15\sqrt{3}.
Substitute x=15x = 15 into TB=x3TB = x\sqrt{3}.

Key Concept

Special Right Triangle Ratios (30∘−60∘−90∘30^\circ-60^\circ-90^\circ)
Estimated Time:1m 30s
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Topics

Pythagorean Theorem and Special Right Triangles

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