To solve the given questions based on the provided information about triangles \( \triangle ABD \) and \( \triangle BCD \), let's analyze each part step by step.
3.1 Find Angle \( B \).
Given:
- \( AD || BC \)
- \( \angle A = 30^\circ \)
Using the properties of parallel lines and alternate interior angles, we know:
\[ \angle A = \angle B \quad \text{(because } AD || BC \text{)} \]
Thus, it follows directly that:
\[ \angle B = 30^\circ \]
So,
\[ \text{Answer: } \angle B = 30^\circ \quad (4 \text{ marks}) \]
3.2 Find Length \( AB \).
To find \( AB \), we can use the information given and the properties of similar triangles. First, we need to find the height of \( \triangle BCD \) from point \( D \) to \( BC \).
Knowing that \( BD = 4 \) and \( DC = 3 \), we can find the length of \( BC \) as:
\[ BC = BD + DC = 4 + 3 = 7 \]
To find \( AB \), we can use the Law of Sines in \( \triangle ABD \):
\[ \frac{AB}{\sin B} = \frac{AD}{\sin D} \]
Since we have \( AD = DC = 3 \) (as they are opposite the same angle), and \( D \) will also equal \( A \) since \( AD \parallel BC \).
Hence,
\[ AB = \frac{AD \cdot \sin B}{\sin A} = \frac{3 \cdot \sin 30^\circ}{\sin 30^\circ} = 3 \]
So
\[ \text{Answer: } AB = 3 \quad (5 \text{ marks}) \]
3.3 Find the Area of \( \triangle BDA \).
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} \]
In triangle \( BDA \), we can choose \( BD \) as the base of the triangle. The height can be calculated as:
\[ \text{Height} = AB \cdot \sin(A) \]
Substituting values, we have:
- \( AB = 3 \)
- \( A = 30^\circ \)
\[ \text{Height} = 3 \cdot \sin(30^\circ) = 3 \cdot \frac{1}{2} = 1.5 \]
Now, calculating the area of triangle \( BDA \):
\[ A = \frac{1}{2} \times BD \times \text{Height} = \frac{1}{2} \times 4 \times 1.5 = 3 \]
So
\[ \text{Answer: Area } = 3 \quad (6 \text{ marks}) \]
Summary of Answers:
- \( \angle B = 30^\circ \) (4 marks)
- \( AB = 3 \) (5 marks)
- Area of triangle \( BDA = 3 \) (6 marks)
Total = 15 marks.
1 answer