The diagram below shows three points, A, B and C on flat ground. AB = 80 m, AC = 100 m and BC = 90 m. C is due east of A.

1. Identify the cosine rule for the required angle

In $\triangle ABC$, $AB=80$, $BC=90$ and $AC=100$.

For $\angle ACB$, the opposite side is $AB$.

Using the cosine rule,
$AB^2=AC^2+BC^2-2(AC)(BC)\cos ACB.$

2. Substitute the side lengths and solve for $\cos ACB$

$80^2=100^2+90^2-2(100)(90)\cos ACB$
$6400=10000+8100-18000\cos ACB$
$18000\cos ACB=11700$
$\cos ACB=\frac{11700}{18000}=0.65.$

3. Calculate the angle and round to one decimal place

$\angle ACB=\cos^{-1}(0.65)$
$\angle ACB=49.458\ldots^\circ$

Therefore,
$\angle ACB=49.5^\circ \text{ correct to 1 decimal place.}$

1. Relate the triangle angle to the direction of C from B

Since $C$ is due east of $A$, the line $CA$ is due west from $C$. From part (a)(i),
$\angle ACB = 49.5^\circ$.

So the line $CB$ is $49.5^\circ$ north of west. Hence the opposite direction, from $B$ to $C$, is $49.5^\circ$ south of east.

2. Convert the direction into a bearing from north

Bearing is measured clockwise from north.

East has bearing $90^\circ$, so
$\text{bearing of } C \text{ from } B = 90^\circ + 49.5^\circ = 139.5^\circ.$

1. Find the angle at D in triangle BDC

Since $A$, $D$ and $C$ lie on a straight line,
$\angle BDC = 180^\circ - \angle BDA$
$\angle BDC = 180^\circ - 70^\circ = 110^\circ$

2. Find the remaining angle in triangle BDC

From part (a)(i), $\angle ACB = 49.5^\circ$.

Since $D$ lies on $AC$, $\angle BCD = 49.5^\circ$.

In triangle $BDC$,
$\angle DBC = 180^\circ - 110^\circ - 49.5^\circ$
$\angle DBC = 20.5^\circ$

3. Apply the sine rule to find CD

Using the sine rule in triangle $BDC$,
$\frac{CD}{\sin 20.5^\circ} = \frac{BC}{\sin 110^\circ}$
$\frac{CD}{\sin 20.5^\circ} = \frac{90}{\sin 110^\circ}$
$CD = \frac{90\sin 20.5^\circ}{\sin 110^\circ}$
$CD = 33.5\text{ m} \;\text{(to 3 s.f.)}$

1. Relate the ratio of the two areas to the ratio of their bases

Since $A$, $D$ and $C$ lie on the same straight line, triangles $ABD$ and $BDC$ have the same perpendicular height from $B$ to $AC$.

Hence,
$\text{Area of }\triangle ABD : \text{Area of }\triangle BDC = AD : DC.$

2. Find the base lengths and form the ratio

From part (a)(iii),
$CD = 33.5\text{ m}.$

Since $AC = 100\text{ m}$,
$AD = 100 - 33.5 = 66.5\text{ m}.$

Therefore,
$AD : DC = 66.5 : 33.5.$

Dividing both parts by $33.5$,
$66.5 : 33.5 = \frac{66.5}{33.5} : 1 = 1.985\ldots : 1.$

Correct to one decimal place,
$1.985\ldots : 1 = 2.0 : 1.$

1. Find the horizontal distance from $M$ to the point vertically below $G$

Since $M$ lies on $AE$, the horizontal projection of $M$ is $A$. The horizontal projection of $G$ is $C$.

In rectangular base $ABCD$,
$AC=\sqrt{AB^2+BC^2}=\sqrt{4^2+3^2}=5\text{ cm}.$

2. Find the vertical height difference between $G$ and $M$

$AE=12\text{ cm}$ and $M$ is the midpoint of $AE$.

So
$AM=\frac{12}{2}=6\text{ cm}.$

Since $G$ is $12\text{ cm}$ above the base, the vertical height of $G$ above $M$ is
$12-6=6\text{ cm}.$

3. Use tangent ratio to find the angle of elevation

Let the angle of elevation of $G$ from $M$ be $\theta$.

$\tan\theta=\frac{\text{vertical height}}{\text{horizontal distance}}=\frac{6}{5}$

$\theta=\tan^{-1}\left(\frac{6}{5}\right)=50.194\ldots^\circ$

$\theta=50.2^\circ$