To solve for the length $h$ from point $B$ to point $D$ in this diagram, we can use trigonometry since we are working with right-angled triangles.

Key Information:

• $AC = 70 \, \text{metres}$
• $\angle DBA = \angle DBC = 90^\circ$ (so triangles $\triangle ABD$ and $\triangle BCD$ are right-angled triangles)
• $\angle DAB = 64^\circ$
• $\angle DCB = 36^\circ$

Steps:

1. Use $\triangle ABD$:

We know that $\angle DAB = 64^\circ$, and $\triangle ABD$ is a right triangle. We can use the tangent function to find $BD$:

$\tan(64^\circ) = \frac{h}{BD}$ $BD = \frac{h}{\tan(64^\circ)}$

2. Use $\triangle BCD$:

In $\triangle BCD$, $\angle DCB = 36^\circ$. We can also use the tangent function here:

$\tan(36^\circ) = \frac{h}{BD}$ From this, we get: $BD = \frac{70}{\tan(36^\circ)}$

3. Solve for $h$:

Since both expressions for $BD$ are equal, we set them equal to each other and solve for $h$:

$\frac{h}{\tan(64^\circ)} = \frac{70}{\tan(36^\circ)}$

Now, I will calculate the value of $h$.The length $h$ from point $B$ to point $D$ is approximately $37.55 \, \text{metres}$.

Would you like more details on the steps or explanations?

Here are five related questions to expand on this concept:

1. How do you apply the sine and cosine rules in non-right-angled triangles?
2. What is the relationship between angles and side lengths in similar triangles?
3. How can you calculate the area of triangles using trigonometry?
4. How would the solution change if angle $\angle DAB$ or $\angle DCB$ were different?
5. How can you use the Pythagorean theorem to solve problems involving right-angled triangles?

Tip: When working with right-angled triangles, remember that trigonometric functions like sine, cosine, and tangent can be used to relate angles to side lengths, making them useful for height and distance problems.