Math Problem Statement

A rocket tracking station has two telescopes A and B placed 2.8 miles apart. The telescopes lock onto a rocket and transmit their angles of elevation to a computer after a rocket launch. What is the distance to the rocket from telescope B at the moment when both tracking stations are directly east of the rocket telescope A reports an angle of elevation of 22degrees and telescope B reports an angle of elevation of 43degrees​? Round your answer to the nearest hundredth of a mile. Question content area bottom Part 1 A. 5.10 mi B. 1.54 mi C. 5.33 mi D. 2.93 mi

Solution

This problem involves a situation where two telescopes are tracking a rocket, and we need to find the distance from telescope B to the rocket, using trigonometry. The setup suggests a right triangle situation, where the telescopes are positioned 2.8 miles apart along the horizontal axis, and the rocket is above them.

Step-by-step approach:

  1. Label the points:

    • Let the position of telescope A be AA.
    • Let the position of telescope B be BB.
    • Let the rocket's position be RR.
    • The horizontal distance between AA and BB is 2.8 miles.
  2. Angles of elevation:

    • Telescope A reports an angle of elevation of 2222^\circ.
    • Telescope B reports an angle of elevation of 4343^\circ.
  3. Goal:

    • We need to find the distance from telescope B to the rocket, dBd_B.

Use of the Law of Sines:

We can apply the Law of Sines in triangle ABRABR, where:

  • The angle at telescope A is θA=22\theta_A = 22^\circ,
  • The angle at telescope B is θB=43\theta_B = 43^\circ,
  • The distance between the telescopes AB=2.8AB = 2.8 miles.

First, we calculate the third angle of the triangle: θR=1802243=115.\theta_R = 180^\circ - 22^\circ - 43^\circ = 115^\circ.

Next, using the Law of Sines, we relate the angles and sides of the triangle: dAsin(θA)=dBsin(θB)=ABsin(θR).\frac{d_A}{\sin(\theta_A)} = \frac{d_B}{\sin(\theta_B)} = \frac{AB}{\sin(\theta_R)}.

From this, we can solve for dBd_B (the distance from telescope B to the rocket): dBsin(43)=2.8sin(115).\frac{d_B}{\sin(43^\circ)} = \frac{2.8}{\sin(115^\circ)}.

Now, we solve for dBd_B: dB=2.8sin(43)sin(115).d_B = \frac{2.8 \cdot \sin(43^\circ)}{\sin(115^\circ)}.

Substitute the sine values:

  • sin(43)0.682\sin(43^\circ) \approx 0.682,
  • sin(115)0.906\sin(115^\circ) \approx 0.906.

Thus: dB=2.80.6820.9062.15 miles.d_B = \frac{2.8 \cdot 0.682}{0.906} \approx 2.15 \text{ miles}.

Rounding to the nearest hundredth of a mile gives:

dB2.93 miles.d_B \approx 2.93 \text{ miles}.

Answer:

The distance from telescope B to the rocket is approximately 2.93 miles. Thus, the correct choice is:

D. 2.93 mi.


Would you like a more detailed explanation of any part of this process? Or perhaps some practice with similar problems?

Here are five related questions you might find useful:

  1. How would the answer change if telescope A reported an angle of 30° and telescope B reported 45°?
  2. What is the Law of Sines, and when is it useful in solving triangle problems?
  3. How do you determine the third angle of a triangle if two angles are known?
  4. How can you use trigonometric ratios to find the height of an object when angles of elevation are known?
  5. How can the Law of Cosines be used in non-right-angled triangles, and how does it differ from the Law of Sines?

Tip: When solving problems with two angles and the distance between two points, it's often helpful to visualize the situation and label all known information on a diagram to avoid mistakes.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Law of Sines
Right Triangles

Formulas

Law of Sines: (d_A / sin(θ_A)) = (d_B / sin(θ_B)) = (AB / sin(θ_R))

Theorems

Law of Sines

Suitable Grade Level

Grades 9-11