QuestionTrigonomet'y Homework
A flagpole $[G H]$, shown in the diagram, is vertical and the ground is inclined at an angle of $5^{\circ}$ to the horizontal between $E$ and $G$. The angles of elevation from $E$ and $F$ to the top of the pole are $35^{\circ}$ and $52^{\circ}$ respectively. The distance from $E$ to $F$ along the incline is 6 m . Find how far $F$ is from the base of the pole $(G)$ along the incline. Give your answer correct to two decimal places.

Studdy Solution
Use the Law of Sines to find the distance from $F$ to $G$.
The triangle $\triangle EFG$ has angles $\angle EFG = 180^\circ - 47^\circ - 30^\circ = 103^\circ$.
Using the Law of Sines:
$$\frac{6}{\sin(103^\circ)} = \frac{x}{\sin(30^\circ)}$$
Solve for $x$:
$$x = \frac{6 \cdot \sin(30^\circ)}{\sin(103^\circ)}$$
Calculate $x$:
$$x \approx \frac{6 \cdot 0.5}{0.97437} \approx 3.08$$
The distance from $F$ to $G$ along the incline is approximately:
$$\boxed{3.08 \text{ meters}}$$