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Difficulty: MediumRight Triangle Trigonometry (SOHCAHTOA)

A kite string of length 5050 meters is attached to a stake anchored in level ground at point KK. The kite is flying at point HH, directly above a landmark LL on the ground, forming right triangle KLHKLH with the right angle at LL. If the angle of elevation from the stake to the kite is θ\theta, such that cos⁡(θ)=2425\cos(\theta) = \frac{24}{25}, what is the vertical height, in meters, of the kite above the ground?

  1. A
    77
  2. 1414Answer
  3. C
    2424
  4. D
    4848
  5. E
    2626

Answer

The vertical height of the kite above the ground is 1414 meters.
The vertical height corresponds to the leg opposite to angle θ\theta. Using cos⁡(θ)=2425\cos(\theta) = \frac{24}{25}, the adjacent side is 4848 meters. Applying the Pythagorean theorem LH=502−482=14LH = \sqrt{50^2 - 48^2} = 14 meters gives the correct vertical height.

Step-by-Step Solution

1
Identify the given trigonometric ratio and side length
Hypotenuse KH=50KH = 50 meters and cos⁡(θ)=adjacenthypotenuse=KL50=2425\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{KL}{50} = \frac{24}{25}.
Cosine relates the adjacent side (ground distance) to the hypotenuse (string length).
2
Calculate the horizontal ground distance KLKL
KL=50×2425=48KL = 50 \times \frac{24}{25} = 48 meters.
Multiply the hypotenuse length by the cosine ratio.
3
Determine the sine ratio or use the Pythagorean theorem to find the vertical height LHLH
Since sin⁡(θ)=1−cos⁡2(θ)=1−(2425)2=725\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{24}{25}\right)^2} = \frac{7}{25}, the height LH=50×sin⁡(θ)=50×725=14LH = 50 \times \sin(\theta) = 50 \times \frac{7}{25} = 14 meters.
The sine ratio relates the opposite vertical height to the hypotenuse.

Key Concept

Right Triangle Trigonometry (SOHCAHTOA) and Pythagorean Triples
Estimated Time:1m 0s
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Topics

Right Triangle Trigonometry (SOHCAHTOA)

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