Answer:
190.20 m
Step-by-step explanation:
Given:
- The bearing of a house from a point A is 319°.
- From a point B, 317 m due east of A, the bearing of the house is 288°.
Let point H be the location of the house.
A bearing is the angle in degrees measured clockwise from north.
Draw a diagram using the given information (see attachment). This forms triangle ABH. Calculate the internal angles of the triangle:
[tex]A = 360^{\circ}-319^{\circ}+90^{\circ} \\\\ A = 131^{\circ}[/tex]
[tex]B = 288^{\circ}-270^{\circ} \\\\ B = 18^{\circ}[/tex]
[tex]A+B+H = 180^{\circ} \\\\131^{\circ}+18^{\circ}+H=\80^{\circ} \\\\149^{\circ}+H=\80^{\circ} \\\\H=31^{\circ}[/tex]
To determine how far the house is from point A, we can use the Law of Sines.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
Let d be the distance between point A and the house.
Therefore, in this case:
- B = 18°
- b = d
- H = 31°
- h = 317
Substitute the values into the formula:
[tex]\dfrac{d}{\sin B}=\dfrac{h}{\sin H} \\\\\\ \dfrac{d}{\sin 18^{\circ}}=\dfrac{317}{\sin 31^{\circ}}[/tex]
Solve for d:
[tex]d=\dfrac{317\sin 18^{\circ}}{\sin 31^{\circ}} \\\\\\ d=190.19639904... \\\\\\ d=190.20\; \sf m\;(nearest\;hundredth)[/tex]
So, the house is 190.20 m from point A.
