Math Problem Statement

The angle of depression from a lookout at the top of a lighthouse (A) down to a boat located at point C is 30°. The boat travels towards the lighthouse and after 1 minute has traveled a distance of 50 m and is now located at point B. The angle of elevation from the boat at B up to the lighthouse lookout is 60°. (a) Write down the value of x° in the diagram. (b) Find the height of the lighthouse. (c) Find the speed of the boat, in meters per second, from C to B.

Solution

Let's solve the given problem step-by-step.

Problem Analysis:

  1. A lighthouse (A) has a lookout at the top. The angle of depression from A to a boat at point C is 30°.
  2. After 1 minute, the boat travels 50 m towards the lighthouse and is now at point B.
  3. The angle of elevation from point B to the lighthouse lookout is 60°.

Given Data:

  • Distance BC=50BC = 50 m.
  • Angle of depression from A to C: AOC=30\angle AOC = 30^\circ.
  • Angle of elevation from B to A: ABO=60\angle ABO = 60^\circ.

Questions:

(a) Write down the value of xx^\circ in the diagram.

(b) Find the height of the lighthouse.

(c) Find the speed of the boat, in meters per second, from C to B.

Solution:

(a) Value of xx^\circ:

The value of xx^\circ is the angle at C, which is 60°. This is because AOC=30\angle AOC = 30^\circ and the angle of elevation at B to A is 60°.

Answer (a): x=60x = 60^\circ.

(b) Height of the lighthouse:

Let the height of the lighthouse be hh. From the right triangle AOBAOB:

Using the tangent of the angle at B: tan(60)=hOB.\tan(60^\circ) = \frac{h}{OB}. Since tan(60)=3\tan(60^\circ) = \sqrt{3}: OB=h3.OB = \frac{h}{\sqrt{3}}.

Now, using the right triangle AOCAOC:

tan(30)=hOC,\tan(30^\circ) = \frac{h}{OC}, and since tan(30)=13\tan(30^\circ) = \frac{1}{\sqrt{3}}: OC=h3.OC = h \cdot \sqrt{3}.

Given that OC=OB+BCOC = OB + BC: h3=h3+50.h \cdot \sqrt{3} = \frac{h}{\sqrt{3}} + 50.

Multiplying the entire equation by 3\sqrt{3}: 3h=h+503.3h = h + 50\sqrt{3}.

Simplify: 2h=503,2h = 50\sqrt{3}, h=25325×1.732=43.3 m.h = 25\sqrt{3} \approx 25 \times 1.732 = 43.3 \text{ m}.

Answer (b): The height of the lighthouse is approximately 43.3 m.

(c) Speed of the boat:

The boat travels from C to B, a distance of 50 m, in 1 minute (60 seconds).

Speed: Speed=DistanceTime=50 m60 s0.833 m/s.\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{50 \text{ m}}{60 \text{ s}} \approx 0.833 \text{ m/s}.

Answer (c): The speed of the boat is approximately 0.833 m/s.


Would you like further details or explanations on any part? Here are some related questions:

  1. How would the height of the lighthouse change if the angles were different?
  2. Can you find the time taken if the boat was moving at a different speed?
  3. What other trigonometric ratios could be used to solve similar problems?
  4. How would you find the horizontal distance covered by the boat if it traveled in a different direction?
  5. How do changes in elevation angles affect distance measurements in trigonometry?

Tip: Always check the units of measurement when calculating speeds, distances, or heights to ensure consistency and accuracy.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Trigonometry
Angle of Depression
Angle of Elevation
Tangent Function

Formulas

tan(θ) = opposite / adjacent
Speed = Distance / Time

Theorems

Trigonometric Ratios in Right-Angled Triangles

Suitable Grade Level

Grade 9-10