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Difficulty: HardAngles of Elevation, Depression, and Bearings

An observer standing at the top of a vertical lighthouse observes two boats, XX and YY, on the surrounding horizontal sea surface. Boat XX lies due South of the lighthouse at an angle of depression of 30∘30^\circ, while boat YY lies due East of the lighthouse at an angle of depression of 45∘45^\circ. If the straight-line distance between boat XX and boat YY is 80 m80\text{ m}, what is the height of the lighthouse?

  1. 40 m40\text{ m}Answer
  2. B
    403 m40\sqrt{3}\text{ m}
  3. C
    402 m40\sqrt{2}\text{ m}
  4. D
    203 m20\sqrt{3}\text{ m}

Answer

The height of the lighthouse is 40 m40\text{ m}.
The height of the lighthouse is 40 m40\text{ m}. Since boat XX is due South and boat YY is due East, the line segments connecting the base of the lighthouse to each boat form a right angle (90∘90^\circ). Using basic trigonometry, the distance to boat XX is h3h\sqrt{3} and to boat YY is hh. Applying Pythagoras' theorem to the right triangle formed on the sea surface gives (h3)2+h2=802(h\sqrt{3})^2 + h^2 = 80^2, which simplifies to 4h2=64004h^2 = 6400, giving h=40 mh = 40\text{ m}.

Step-by-Step Solution

1
Express the horizontal distance from the lighthouse base LL to boat XX in terms of height hh.
LX=htan⁡30∘=h3 mLX = \frac{h}{\tan 30^\circ} = h\sqrt{3}\text{ m}
The angle of elevation from boat XX to the top of the lighthouse is equal to the angle of depression (30∘30^\circ).
2
Express the horizontal distance from the lighthouse base LL to boat YY in terms of height hh.
LY=htan⁡45∘=h mLY = \frac{h}{\tan 45^\circ} = h\text{ m}
The angle of elevation from boat YY to the top of the lighthouse is 45∘45^\circ.
3
Set up Pythagoras' theorem for right-angled triangle XLYXLY on the horizontal plane.
XY2=LX2+LY2  ⟹  802=(h3)2+h2XY^2 = LX^2 + LY^2 \implies 80^2 = (h\sqrt{3})^2 + h^2
Boat XX is due South and boat YY is due East of the lighthouse base, making ∠XLY=90∘\angle XLY = 90^\circ.
4
Solve the algebraic equation for hh.
6400=3h2+h2=4h2  ⟹  h2=1600  ⟹  h=40 m6400 = 3h^2 + h^2 = 4h^2 \implies h^2 = 1600 \implies h = 40\text{ m}
Dividing 64006400 by 44 gives 16001600, whose square root is 4040.

Key Concept

3D Geometry combining Angles of Elevation/Depression with Perpendicular Bearings
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Topics

Angles of Elevation, Depression, and Bearings

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