The problem involves a triangle ABC, where angle A is $115^{\circ}$, side a (opposite to angle A) is 65m, and side b (opposite to angle B) is 32m. The task is to find the value of angle B. The sine rule is used to relate the sides and angles of the triangle.

1. Problem Description

The problem involves a triangle ABC, where angle A is $115^{\circ}$, side a (opposite to angle A) is 65m, and side b (opposite to angle B) is 32m. The task is to find the value of angle B. The sine rule is used to relate the sides and angles of the triangle.

2. Solution Steps

The sine rule is given by:

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Given $a = 65$, $b = 32$, and $A = 115^{\circ}$, we can use the sine rule to find $\sin B$:

$\frac{65}{\sin 115^{\circ}} = \frac{32}{\sin B}$

Cross-multiply to get:

$65 \sin B = 32 \sin 115^{\circ}$

Divide both sides by 65:

$\sin B = \frac{32 \sin 115^{\circ}}{65}$

Now, calculate the value of $\sin B$:

$\sin B = \frac{32 \times 0.9063}{65} \approx \frac{29.0016}{65} \approx 0.44617$

To find the angle B, take the inverse sine of 0.44617:

$B = \arcsin(0.44617) \approx 26.49^{\circ}$ (or 26.5 if rounded)

The image also mentions the answer as 26.

5.

3. Final Answer

The angle B is approximately $26.5^{\circ}$.