Examkin

Question

Difficulty: Very hardLaw of Sines and Law of Cosines

A surveyor stands at point AA on flat ground and measures the angle of elevation to the top of a vertical tower, TT, at point CC to be 30∘30^\circ. Another surveyor at point BB, which is 100100 meters away from AA on the same flat ground, measures ∠CAB=40∘\angle CAB = 40^\circ and ∠CBA=65∘\angle CBA = 65^\circ. Which of the following expressions represents the height, in meters, of the tower?

  1. 100sin⁡(65∘)tan⁡(30∘)sin⁡(75∘)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}Answer
  2. B
    100sin⁡(40∘)tan⁡(30∘)sin⁡(75∘)\frac{100 \sin(40^\circ) \tan(30^\circ)}{\sin(75^\circ)}
  3. C
    100sin⁡(65∘)sin⁡(30∘)sin⁡(75∘)\frac{100 \sin(65^\circ) \sin(30^\circ)}{\sin(75^\circ)}
  4. D
    100sin⁡(75∘)tan⁡(30∘)sin⁡(65∘)\frac{100 \sin(75^\circ) \tan(30^\circ)}{\sin(65^\circ)}
  5. E
    100sin⁡(65∘)tan⁡(30∘)sin⁡(105∘)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(105^\circ)}

Answer

The correct expression is 100sin⁡(65∘)tan⁡(30∘)sin⁡(75∘)\frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
The correct expression is derived by first applying the Law of Sines to find the length of the ground segment ACAC, and then using right-triangle trigonometry to determine the height of the tower. In triangle ABCABC, the third angle ∠ACB\angle ACB is 180∘−(40∘+65∘)=75∘180^\circ - (40^\circ + 65^\circ) = 75^\circ. The Law of Sines gives ACsin⁡(65∘)=100sin⁡(75∘)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)}, which simplifies to AC=100sin⁡(65∘)sin⁡(75∘)AC = \frac{100\sin(65^\circ)}{\sin(75^\circ)}. Since the tower is vertical, triangle ACTACT is a right triangle with tan⁡(30∘)=hAC\tan(30^\circ) = \frac{h}{AC}. Substituting ACAC yields h=100sin⁡(65∘)tan⁡(30∘)sin⁡(75∘)h = \frac{100\sin(65^\circ)\tan(30^\circ)}{\sin(75^\circ)}.

Step-by-Step Solution

1
Calculate the measure of the third angle in the ground triangle △ABC\triangle ABC.
∠ACB=180∘−(40∘+65∘)=75∘\angle ACB = 180^\circ - (40^\circ + 65^\circ) = 75^\circ
The sum of angles in any triangle must equal 180∘180^\circ.
2
Apply the Law of Sines to find the distance from point AA to the base of the tower at point CC (ACAC).
ACsin⁡(65∘)=100sin⁡(75∘)⇒AC=100sin⁡(65∘)sin⁡(75∘)\frac{AC}{\sin(65^\circ)} = \frac{100}{\sin(75^\circ)} \Rightarrow AC = \frac{100 \sin(65^\circ)}{\sin(75^\circ)}
The Law of Sines states that the ratio of a side length to the sine of its opposite angle is constant in a triangle.
3
Use the right-triangle trigonometric ratio for the vertical tower height hh from point AA.
tan⁡(30∘)=hAC⇒h=ACtan⁡(30∘)\tan(30^\circ) = \frac{h}{AC} \Rightarrow h = AC \tan(30^\circ)
In right triangle △ACT\triangle ACT, the tangent of the angle of elevation is the ratio of the opposite side (height hh) to the adjacent side (ACAC).
4
Substitute the expression for ACAC into the equation for hh.
h=100sin⁡(65∘)tan⁡(30∘)sin⁡(75∘)h = \frac{100 \sin(65^\circ) \tan(30^\circ)}{\sin(75^\circ)}
Replacing ACAC with its equivalent algebraic expression yields the final height in terms of the given parameters.

Key Concept

Applying the Law of Sines to find a missing side length in a non-right triangle and then using right-triangle trigonometric ratios to solve a 3D geometry problem.
Estimated Time:2m 0s
Rate this question

Topics

Law of Sines and Law of Cosines

More questions from this topic