Question

three points a, b and c are on level ground. the bearing of b from a is 300° and b is due north of c. given that ab = 800m and ac = 700m, (i) show that the bearing of a from b is 120°. (ii) find the possible bearing(s) of a from c. (iii) calculate the length of bc, given further that the bearing of a from c is less than 090°.

Explanation:

Step1: Understand bearing concept

Bearings are measured clock - wise from the north direction. If the bearing of $B$ from $A$ is $300^{\circ}$, then the angle between the line $AB$ and the north - south line through $A$ is $360^{\circ}-300^{\circ}=60^{\circ}$. So the angle $\angle BAN = 60^{\circ}$ (where $N$ is the north direction). The bearing of $A$ from $B$ is $180^{\circ}-60^{\circ}=120^{\circ}$.

Step2: Use the cosine - law to find $\cos\angle BAC$

The cosine - law states that $BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos\angle BAC$. First, we use the sine - law in $\triangle ABC$ to find $\angle ABC$ or $\angle ACB$. By the sine - law, $\frac{\sin\angle ACB}{AB}=\frac{\sin\angle ABC}{AC}=\frac{\sin\angle BAC}{BC}$. In $\triangle ABC$, let $\angle BAC=\alpha$. We know $AB = 800$ m and $AC = 700$ m.
We use the sine - law $\frac{\sin\angle ACB}{800}=\frac{\sin\angle ABC}{700}$. Also, from the bearing information, we know the relationship between the angles. Since $B$ is due north of $C$, we consider the geometric relationship of the triangle on the horizontal plane.
The sine - law: $\frac{\sin\angle ACB}{800}=\frac{\sin\angle ABC}{700}$. Let's find $\angle ABC$ first. We know that the bearing of $A$ from $B$ is $120^{\circ}$, so in $\triangle ABC$, if we consider the angle at $B$, we can find the possible values of the bearing of $A$ from $C$.
We know that $\frac{\sin\angle ACB}{800}=\frac{\sin\angle ABC}{700}$. Let $\theta$ be the angle between $AC$ and the north - south line through $C$.
We use the sine - law $\sin\angle ACB=\frac{800\sin\angle ABC}{700}$.
Since the sum of angles in a triangle is $180^{\circ}$, and we know the relationship between the bearings and the angles of the triangle.
If we consider the bearing of $A$ from $C$, we first find $\angle ACB$.
By the sine - law $\frac{\sin\angle ACB}{800}=\frac{\sin120^{\circ}}{700}$, so $\sin\angle ACB=\frac{800\sin120^{\circ}}{700}=\frac{800\times\frac{\sqrt{3}}{2}}{700}=\frac{4\sqrt{3}}{7}\approx0.9897$. So $\angle ACB\approx81.8^{\circ}$ or $\angle ACB = 180^{\circ}-81.8^{\circ}=98.2^{\circ}$.
The bearing of $A$ from $C$:
Case 1: If $\angle ACB\approx81.8^{\circ}$, and $B$ is due north of $C$, the bearing of $A$ from $C$ is $360^{\circ}-(90^{\circ}-81.8^{\circ})=351.8^{\circ}\approx352^{\circ}$
Case 2: If $\angle ACB = 98.2^{\circ}$, the bearing of $A$ from $C$ is $90^{\circ}+ 8.2^{\circ}=98.2^{\circ}\approx98^{\circ}$

Step3: Use the cosine - law to find $BC$ when bearing of $A$ from $C$ is less than $90^{\circ}$

If the bearing of $A$ from $C$ is less than $90^{\circ}$, then $\angle ACB$ is such that we use the cosine - law $BC^{2}=AB^{2}+AC^{2}-2\cdot AB\cdot AC\cdot\cos\angle BAC$.
Since the bearing of $A$ from $C$ is less than $90^{\circ}$, $\angle ACB$ is the acute angle. We know $AB = 800$ m, $AC = 700$ m and $\angle BAC$ can be found from the sine - law results.
Using the cosine - law $BC^{2}=800^{2}+700^{2}-2\times800\times700\times\cos\angle BAC$.
First, from the sine - law $\frac{\sin\angle ACB}{800}=\frac{\sin120^{\circ}}{700}$, we find $\angle ACB\approx81.8^{\circ}$, then $\angle BAC=180^{\circ}-120^{\circ}-81.8^{\circ}+180^{\circ}=38.2^{\circ}$
$BC^{2}=800^{2}+700^{2}-2\times800\times700\times\cos38.2^{\circ}$
$BC^{2}=640000 + 490000-1120000\times0.785$
$BC^{2}=640000+490000 - 879200$
$BC^{2}=250800$
$BC=\sqrt{250800}\approx500.8\approx501$ m

Answer:

(i) Shown as above.
(ii) $98^{\circ}$ or $352^{\circ}$
(iii) $501$ m