Question

determine the length of x, to the nearest tenth of a metre.
a) 115.6 m
b) 123.4 m
c) 202.2 m
d) 103.0 m

Explanation:

Step1: Use the Law of Sines

For the first triangle with angle \(27^{\circ}\) and side \(225m\) (opposite to \(51^{\circ}\) in the second - like part).
Let's first consider the two triangles. In a triangle, by the Law of Sines \(\frac{a}{\sin A}=\frac{b}{\sin B}\).
For the non - right - angled part: assume we have two triangles. Let's consider the relationship between the sides and angles.
We know that in a triangle, if we have two angles and a side, we can find other sides.
Let's assume we have two triangles combined. Let's first find the side related to the common side.
Let's use the Law of Sines formula \(\frac{x}{\sin65^{\circ}}=\frac{225}{\sin(180-(51 + 27))^{\circ}}\)
Since the sum of angles in a triangle is \(180^{\circ}\), the angle opposite to the side of length \(225m\) (in the combined - triangle consideration) is \(180-(51 + 27)=102^{\circ}\)
So, from \(\frac{x}{\sin65^{\circ}}=\frac{225}{\sin102^{\circ}}\)

Step2: Solve for \(x\)

Cross - multiply: \(x=\frac{225\times\sin65^{\circ}}{\sin102^{\circ}}\)
We know that \(\sin65^{\circ}\approx0.9063\) and \(\sin102^{\circ}\approx0.9781\)
\(x=\frac{225\times0.9063}{0.9781}=\frac{203.9175}{0.9781}\approx202.2\)

Answer:

c) \(202.2m\)