Geometry and Trigonometry II

Module 2 handled right-angled triangles. Here the right angle disappears, and two new rules take over. Alongside them sit the circle theorems, where naming the theorem is worth as much as the number.

Circle theorems

Seven results, each needing to be quoted by name.

1. The angle at the centre is twice the angle at the circumference, when both stand on the same arc. An angle of $130°$ at the centre gives $65°$ at the circumference.

2. The angle in a semicircle is $90°$. Any triangle drawn on a diameter, with its third point on the circumference, is right-angled. This is really theorem 1 with a centre angle of $180°$.

3. Angles in the same segment are equal. Two angles at the circumference standing on the same arc are equal, wherever on that arc they sit.

4. Opposite angles of a cyclic quadrilateral add to $180°$. A cyclic quadrilateral has all four vertices on the circle. If one angle is $105°$, the one opposite is $75°$.

5. A tangent is perpendicular to the radius at the point of contact — always exactly $90°$.

6. Tangents from an external point are equal in length. The two tangents from a point outside the circle make an isosceles triangle with the chord joining their contact points.

7. Alternate segment theorem. The angle between a tangent and a chord equals the angle in the alternate segment — the angle at the circumference on the other side of the chord.

Angles of elevation and depression

Both are measured from the horizontal.

• The angle of elevation of an object is the angle you look up through from the horizontal.
• The angle of depression is the angle you look down through from the horizontal.

A cliff is 40 m high and a boat is 60 m from its base. Find the angle of elevation of the cliff top from the boat.

Opposite 40, adjacent 60, so use tangent:

$$\tan\theta = \frac{40}{60} \quad \Rightarrow \quad \theta = 33.7° \text{ (1 d.p.)}$$

Bearings

A bearing is measured:

• from north,
• clockwise,
• and written with three figures.

So due east is $090°$, due south is $180°$, due west is $270°$, and an angle of $70°$ is written $070°$.

The bearing of $B$ from $A$ is $070°$. Find the bearing of $A$ from $B$.

The two north lines are parallel, so the back bearing differs by $180°$:

$$070° + 180° = 250°$$

The sine rule

For any triangle,

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$

where side $a$ is opposite angle $A$, and so on.

Use the sine rule when you have a matching pair — a side and the angle opposite it — plus one more piece of information.

In triangle $ABC$, $A = 40°$, $B = 75°$ and $a = 8$ cm. Find $b$.

$$\frac{b}{\sin 75°} = \frac{8}{\sin 40°}$$ $$b = \frac{8 \sin 75°}{\sin 40°} = 12.0 \text{ cm (3 s.f.)}$$

The cosine rule

$$a^2 = b^2 + c^2 - 2bc\cos A$$

Use it when the sine rule cannot start — that is, when you have:

• two sides and the angle between them, and want the third side; or
• all three sides, and want an angle.

$b = 7$ cm, $c = 9$ cm and $A = 60°$. Find $a$.

$$a^2 = 7^2 + 9^2 - 2(7)(9)\cos 60°$$ $$= 49 + 81 - 126(0.5) = 130 - 63 = 67$$ $$a = \sqrt{67} = 8.19 \text{ cm (3 s.f.)}$$

For an angle, rearrange:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

A triangle has sides 5, 6 and 7 cm. Find the largest angle.

The largest angle faces the longest side, so take $a = 7$:

$$\cos A = \frac{25 + 36 - 49}{2(5)(6)} = \frac{12}{60} = 0.2$$ $$A = \cos^{-1}(0.2) = 78.5° \text{ (1 d.p.)}$$

Area of a triangle

When you know two sides and the angle between them,

$$\text{Area} = \frac12 ab\sin C$$

Two sides are 6 cm and 8 cm with an angle of $50°$ between them.

$$\text{Area} = \frac12 (6)(8)\sin 50° = 24 \sin 50° = 18.4 \text{ cm}^2 \text{ (3 s.f.)}$$

Choosing between the three