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

Let's work through the problem step by step.

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

From the diagram, \( BQ \) is the side opposite the angle \( 36^\circ \), and \( PQ = 50 \) meters is the hypotenuse. Using the sine rule:

\[
\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}}
\]

So, applying it to this triangle:

\[
\sin 36^\circ = \frac{BQ}{50}
\]

Multiplying both sides by 50:

\[
BQ = 50 \sin 36^\circ
\]

This shows the result for part (a).

### b) Find the height \( h \) of the tree

To find the height \( h \), we use the right triangle \( \triangle BTQ \) where \( h \) is the opposite side to the angle \( 28^\circ \), and \( BQ \) is the adjacent side.

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

\[
BQ = 50 \times \sin 36^\circ \approx 29.39 \, \text{meters}
\]

Now, using the tangent function:

\[
\tan 28^\circ = \frac{h}{BQ}
\]

Solving for \( h \):

\[
h = BQ \times \tan 28^\circ = 29.39 \times \tan 28^\circ
\]

\[
h \approx 29.39 \times 0.5317 \approx 15.63 \, \text{meters}
\]

Rounding to the nearest meter, \( h \approx 16 \, \text{meters} \).

### c) Find the angle of elevation of \( T \) from \( P \)

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

First, calculate \( BP \):

\[
BP = 50 \times \cos 36^\circ \approx 50 \times 0.809 \approx 40.45 \, \text{meters}
\]

Now, using the tangent function again:

\[
\tan \theta = \frac{h}{BP} = \frac{16}{40.45}
\]

\[
\tan \theta \approx 0.3956
\]

Taking the inverse tangent:

\[
\theta = \tan^{-1}(0.3956) \approx 21^\circ
\]

Thus, the angle of elevation of \( T \) from \( P \) is approximately \( 21^\circ \).

---

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.

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.

Solve a trigonometry problem involving the height of a tree, angles of elevation, and bearings. Learn to apply the sine and tangent rules to find the height of a tree and angle of elevation, suitable for Grades 10-12.

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