Math Problem Statement

answers are 8a) Answers will vary b) 16m c) 21°

A tree BT is due north of an observer at P and due west of an observer at Q. The two observers are 50m apart and the bearing of Q from P is 36°. The angle of elevation of T from Q is 28°. a) Show that BQ = 50 sin 36°. b) Find the height h of the tree, correct to the nearest meter. c) Find the angle of elevation of T from P, correct to the nearest degree.

Solution

Let's work through the problem step by step.

a) Show that BQ=50sin36BQ = 50 \sin 36^\circ

From the diagram, BQBQ is the side opposite the angle 3636^\circ, and PQ=50PQ = 50 meters is the hypotenuse. Using the sine rule:

sinθ=oppositehypotenuse\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}

So, applying it to this triangle:

sin36=BQ50\sin 36^\circ = \frac{BQ}{50}

Multiplying both sides by 50:

BQ=50sin36BQ = 50 \sin 36^\circ

This shows the result for part (a).

b) Find the height hh of the tree

To find the height hh, we use the right triangle BTQ\triangle BTQ where hh is the opposite side to the angle 2828^\circ, and BQBQ is the adjacent side.

We already know that BQ=50sin36BQ = 50 \sin 36^\circ, so:

BQ=50×sin3629.39metersBQ = 50 \times \sin 36^\circ \approx 29.39 \, \text{meters}

Now, using the tangent function:

tan28=hBQ\tan 28^\circ = \frac{h}{BQ}

Solving for hh:

h=BQ×tan28=29.39×tan28h = BQ \times \tan 28^\circ = 29.39 \times \tan 28^\circ

h29.39×0.531715.63metersh \approx 29.39 \times 0.5317 \approx 15.63 \, \text{meters}

Rounding to the nearest meter, h16metersh \approx 16 \, \text{meters}.

c) Find the angle of elevation of TT from PP

For this part, we will use the triangle BPT\triangle BPT. We need to find the angle of elevation of TT from PP, denoted as θ\theta. In this case, we know the opposite side h16h \approx 16 meters, and the adjacent side is 50cos3650 \cos 36^\circ (the distance BPBP).

First, calculate BPBP:

BP=50×cos3650×0.80940.45metersBP = 50 \times \cos 36^\circ \approx 50 \times 0.809 \approx 40.45 \, \text{meters}

Now, using the tangent function again:

tanθ=hBP=1640.45\tan \theta = \frac{h}{BP} = \frac{16}{40.45}

tanθ0.3956\tan \theta \approx 0.3956

Taking the inverse tangent:

θ=tan1(0.3956)21\theta = \tan^{-1}(0.3956) \approx 21^\circ

Thus, the angle of elevation of TT from PP is approximately 2121^\circ.


Let me know if you would like further clarification! Here are 5 questions related to this:

  1. How do we calculate the sine and cosine of angles?
  2. What is the relationship between angle of elevation and height?
  3. Why did we use the tangent function to find the height in part b?
  4. How is the hypotenuse related to the adjacent side and angle in right-angled triangles?
  5. What role do bearings play in solving trigonometric problems like this one?

Tip: In solving trigonometric problems, understanding the right triangle relationships is key. Always identify which sides (opposite, adjacent, hypotenuse) correspond to the angles provided.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Angle of Elevation
Bearings

Formulas

sin θ = opposite/hypotenuse
tan θ = opposite/adjacent
cos θ = adjacent/hypotenuse

Theorems

Sine Rule
Tangent Function in Right Triangles

Suitable Grade Level

Grades 10-12