https://file.aiquickdraw.com/mathbot/file/1728641451122q79ois3q.jpg

answers are 
7a) BQ = 30tan 72° 
b) 145m

Let's work through the questions for the new problem.

### Question 7:
You are given a vertical tower \( BT \) of height 30 m, with two landmarks \( P \) and \( Q \) on level ground. The angles of elevation from \( P \) and \( Q \) to the top of the tower \( T \) are 15° and 18°, respectively.

### a) Show that \( BP = 30 \tan 75^\circ \), and find a similar expression for \( BQ \).

#### Part a) \( BP = 30 \tan 75^\circ \)

From the diagram, consider the right-angled triangle \( \triangle BPT \), where:
- The height of the tower \( BT = 30 \, \text{m} \).
- The angle of elevation from \( P \) to \( T \) is 15°, meaning the angle \( \angle BTP = 15^\circ \).

The angle between \( BP \) and the horizontal at point \( B \) is \( 90^\circ - 15^\circ = 75^\circ \).

In triangle \( \triangle BPT \), we can use the tangent function:
\[
\tan 75^\circ = \frac{BT}{BP}.
\]
Rearranging for \( BP \):
\[
BP = \frac{BT}{\tan 75^\circ}.
\]
Substitute \( BT = 30 \, \text{m} \):
\[
BP = \frac{30}{\tan 75^\circ}.
\]
Using the reciprocal property of tangent:
\[
BP = 30 \tan 75^\circ.
\]

#### Similar expression for \( BQ \)

Similarly, in triangle \( \triangle BQT \), the angle of elevation from \( Q \) to \( T \) is 18°, meaning the angle \( \angle BTQ = 18^\circ \).

Thus, the angle between \( BQ \) and the horizontal at point \( B \) is \( 90^\circ - 18^\circ = 72^\circ \).

Again, using the tangent function:
\[
\tan 72^\circ = \frac{BT}{BQ}.
\]
Rearranging for \( BQ \):
\[
BQ = \frac{BT}{\tan 72^\circ} = 30 \tan 72^\circ.
\]

### b) Find, correct to the nearest metre, the distance between the two landmarks.

To find the distance \( PQ \), we apply Pythagoras' theorem to the right-angled triangle \( \triangle BPQ \), since \( P \) is due south of \( B \) and \( Q \) is due east of \( B \).

From part a), we have the expressions for \( BP \) and \( BQ \):
\[
BP = 30 \tan 75^\circ \quad \text{and} \quad BQ = 30 \tan 72^\circ.
\]
We can compute these values:
- \( \tan 75^\circ \approx 3.732 \)
- \( \tan 72^\circ \approx 3.078 \)

Substitute these into the equations:
\[
BP \approx 30 \times 3.732 = 111.96 \, \text{m}.
\]
\[
BQ \approx 30 \times 3.078 = 92.34 \, \text{m}.
\]

Now, using Pythagoras' theorem in \( \triangle BPQ \):
\[
PQ^2 = BP^2 + BQ^2.
\]
Substitute the values for \( BP \) and \( BQ \):
\[
PQ^2 = (111.96)^2 + (92.34)^2 = 12511.84 + 8538.28 = 21050.12.
\]
Taking the square root:
\[
PQ \approx \sqrt{21050.12} \approx 145.06 \, \text{m}.
\]
Rounding to the nearest metre:
\[
PQ \approx 145 \, \text{m}.
\]

---

Would you like further clarification on any of these steps or additional explanations?

### Related Questions:
1. How can tangent be used to find distances in other geometric problems?
2. What would change if the angles of elevation were measured from different directions?
3. How is Pythagoras' theorem applied in 3D settings?
4. Can you derive the same solution using cosine or sine rules instead?
5. How do the values of trigonometric functions like tangent change as the angle approaches 90°?

#### Tip:
When using trigonometric functions, it helps to visualize the situation in terms of right triangles to quickly apply formulas like tangent and sine.

Two landmarks P and Q on level ground are observed from the top T of a vertical tower BT of height 30m. Landmark P is due south of the tower, while landmark Q is due east of the tower. The angles of elevation of T from P and Q are 15° and 18° respectively. (a) Show that BP = 30tan 75°, and find a similar expression for BQ. (b) Find, correct to the nearest metre, the distance between the two landmarks.

Solve a trigonometry problem involving a vertical tower, landmarks, and angles of elevation. Learn how to express BP and BQ using tangent and calculate the distance between two landmarks using Pythagoras' theorem. Suitable for Grade 10-12 students.

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