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Difficulty: Very hardRight Triangle Trigonometry (SOHCAHTOA)

An observer stands at point PP on horizontal ground and measures the angle of elevation to the top of a vertical tower, TT, as θ\theta. The observer then walks a distance of dd meters directly toward the base of the tower to point QQ. From point QQ, the angle of elevation to the top of the tower is 2θ2\theta. If cos⁡(θ)=45\cos(\theta) = \frac{4}{5}, what is the ratio of the height of the tower to the distance dd?

  1. A
    35\frac{3}{5}
  2. B
    34\frac{3}{4}
  3. C
    1225\frac{12}{25}
  4. 2425\frac{24}{25}Answer
  5. E
    725\frac{7}{25}

Answer

2425\frac{24}{25}
The correct answer of 2425\frac{24}{25} is found by first identifying that the triangle formed by the tower's top and the two observer positions is isosceles. Since the exterior angle is 2θ2\theta and one interior angle is θ\theta, the other interior angle must also be θ\theta, making the side lengths opposite these angles equal (dd). By dropping an altitude inside this isosceles triangle, we form two right triangles, allowing us to find the hypotenuse of the larger right triangle as 2dcos⁡(θ)=1.6d2d\cos(\theta) = 1.6d. Applying the sine definition to the larger right triangle yields sin⁡(θ)=h1.6d\sin(\theta) = \frac{h}{1.6d}. Since cos⁡(θ)=45\cos(\theta) = \frac{4}{5}, we have sin⁡(θ)=35\sin(\theta) = \frac{3}{5}. Solving for hd\frac{h}{d} gives hd=1.6×35=2425\frac{h}{d} = 1.6 \times \frac{3}{5} = \frac{24}{25}.

Step-by-Step Solution

1
Analyze the angles in the triangle formed by the top of the tower TT, the first position PP, and the second position QQ.
The exterior angle at vertex QQ is 2θ2\theta, and the opposite interior angle at vertex PP is θ\theta. Therefore, the third angle ∠PTQ\angle PTQ is 2θ−θ=θ2\theta - \theta = \theta. Since two angles are equal, triangle PTQPTQ is isosceles with QT=PQ=dQT = PQ = d.
This identifies the length of segment QTQT in terms of the walking distance dd.
2
Drop a perpendicular altitude from QQ to segment PTPT meeting at point MM. Use right triangle trigonometry in the resulting right triangle PMQPMQ.
In right triangle PMQPMQ, the hypotenuse is PQ=dPQ = d and the angle is θ\theta. Thus, the adjacent side is PM=dcos⁡(θ)=45d=0.8dPM = d \cos(\theta) = \frac{4}{5}d = 0.8d. Because the altitude of an isosceles triangle bisects the base, the total length PT=2×PM=85d=1.6dPT = 2 \times PM = \frac{8}{5}d = 1.6d.
This determines the length of the hypotenuse PTPT of the large right triangle PRTPRT in terms of dd.
3
Apply the sine ratio to the large right triangle PRTPRT to find the ratio of the height hh to the distance dd.
In right triangle PRTPRT, the angle at PP is θ\theta, the opposite side is hh (height of the tower), and the hypotenuse is PT=1.6dPT = 1.6d. Therefore, sin⁡(θ)=hPT=h1.6d\sin(\theta) = \frac{h}{PT} = \frac{h}{1.6d}. Since cos⁡(θ)=45\cos(\theta) = \frac{4}{5}, we have sin⁡(θ)=35\sin(\theta) = \frac{3}{5}. Substituting this gives 35=h1.6d  ⟹  hd=1.6×35=2425\frac{3}{5} = \frac{h}{1.6d} \implies \frac{h}{d} = 1.6 \times \frac{3}{5} = \frac{24}{25}.
This solves for the ratio of the tower height to the distance dd.

Key Concept

Applying SOHCAHTOA and geometric properties of triangles to solve multi-step trigonometry problems
Estimated Time:3m 0s
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Topics

Right Triangle Trigonometry (SOHCAHTOA)

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