Chapter 5: Review

The chapter problem, solved

A coastguard station $S$ and a lighthouse $L$ stand $10$ km apart. A beacon $B$ is at sea, with angle $30^\circ$ at $S$ and angle $105^\circ$ at $L$ . The harbour at $S$ is a sector of radius $4$ km and angle $0.9$ radians. The tide height is $h = 12\cos\theta + 5\sin\theta$ metres.

Step 1. Find the angle at the beacon. The angles of a triangle add to $180^\circ$ , so

\angle B = 180^\circ - 30^\circ - 105^\circ = 45^\circ.

Step 2. Find the two distances with the sine rule. The side $LB$ is opposite the $30^\circ$ angle and $SL = 10$ km is opposite the $45^\circ$ angle:

\frac{LB}{\sin 30^\circ} = \frac{10}{\sin 45^\circ}, \qquad LB = \frac{10 \times \frac{1}{2}}{\frac{\sqrt{2}}{2}} = 5\sqrt{2} \approx 7.07 \text{ km}.

The side $SB$ is opposite the $105^\circ$ angle. In Section 5.8 you proved that $\sin 105^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ . Therefore

SB = \frac{10\sin 105^\circ}{\sin 45^\circ} = 10 \times \frac{\sqrt{6} + \sqrt{2}}{4} \times \frac{2}{\sqrt{2}} = 5(\sqrt{3} + 1) \approx 13.7 \text{ km}.

The beacon is about $7.07$ km from the lighthouse and about $13.7$ km from the station.

Step 3. Find the area of the triangle. Use two sides and the included angle at $S$ :

\text{area} = \tfrac{1}{2} \times 10 \times 5(\sqrt{3} + 1) \times \sin 30^\circ = \tfrac{1}{2} \times 10 \times 5(\sqrt{3} + 1) \times \tfrac{1}{2} = \frac{25(\sqrt{3} + 1)}{2} \approx 34.2 \text{ km}^2.

Step 4. Find the harbour measurements. The angle is already in radians, so

\text{arc} = r\theta = 4 \times 0.9 = 3.6 \text{ km},

\text{area} = \tfrac{1}{2}r^2\theta = \tfrac{1}{2} \times 16 \times 0.9 = 7.2 \text{ km}^2.

The curved sea wall is $3.6$ km long and the harbour holds $7.2$ km $^2$ of water.

Step 5. Rewrite the tide model as one wave. Here $a = 12$ and $b = 5$ , so

R = \sqrt{12^2 + 5^2} = 13, \qquad \tan\alpha = \frac{5}{12}, \qquad \alpha \approx 0.395 \text{ radians}.

Both coefficients are positive, so $\alpha$ is acute and

h = 13\cos(\theta - 0.395).

The greatest height is $R = 13$ m. The water is never $17$ m high, because the two terms do not peak together.

Step 6. Find when the gate closes. The gate closes when $h = 10$ :

13\cos(\theta - 0.395) = 10, \qquad \cos(\theta - 0.395) = \frac{10}{13} \approx 0.7692.

The principal value is $\cos^{-1}(0.7692) \approx 0.693$ radians, and cosine takes the same value with the opposite sign of the angle. So

\theta - 0.395 \approx 0.693 \quad \text{or} \quad \theta - 0.395 \approx -0.693.

\theta \approx 1.088 \quad \text{or} \quad \theta \approx -0.298.

The tide turns through a full circle of $2\pi$ radians every $12$ hours, so the time in hours is $t = \dfrac{12\theta}{2\pi} = \dfrac{6\theta}{\pi}$ .

t \approx \frac{6 \times 1.088}{\pi} \approx 2.08 \text{ h}.

The second angle is negative, so add $2\pi$ : $\theta \approx 5.985$ , giving

t \approx \frac{6 \times 5.985}{\pi} \approx 11.4 \text{ h}.

The chapter's answer. The beacon is about $7.07$ km from the lighthouse and about $13.7$ km from the station, and the triangle has area about $34.2$ km $^2$ . The harbour sea wall is $3.6$ km long and the water covers $7.2$ km $^2$ . The tide reaches a greatest height of $13$ m and passes $10$ m at about $2.08$ hours and again at about $11.4$ hours.

Mixed practice — choose the method first

For each question, say which method you will use and why before you calculate. The items mix this chapter with Chapters 1, 2 and 3, so the choice is not always obvious.

Work out the exact value of $\tan\dfrac{4\pi}{3}$ .
A triangle has sides $7$ cm, $8$ cm and $11$ cm. Find the largest angle.
A sector has radius $6$ cm and angle $1.4$ radians. Find the arc length and the sector area.
Prove that $\dfrac{\cos\theta}{1 - \sin\theta} \equiv \dfrac{1 + \sin\theta}{\cos\theta}$ for $\cos\theta \neq 0$ .
Write $8\cos\theta - 6\sin\theta$ in the form $R\cos(\theta - \alpha)$ , and state its greatest value.
Solve $4\sin\theta = 3$ for $0 \le \theta < 2\pi$ , giving answers in radians to three decimal places.
Expand $(2x - 3)^3$ .
Find the equation of the line through $(3, -1)$ perpendicular to $y = 2x + 5$ .
Show that the sum of two odd numbers is even.
Find where the curve $y = x^2 - 4x + 3$ crosses the $x$ -axis.
A wave is modelled by $y = 6\sin(3t) + 1$ . State its amplitude, period and range.
A force of $30$ N acts at $55^\circ$ above the horizontal. Find its two perpendicular components, to three significant figures.

Answers

Exact value, third quadrant. $\dfrac{4\pi}{3}$ is $240^\circ$ , where tangent is positive. The related acute angle is $\dfrac{\pi}{3}$ , so $\tan\dfrac{4\pi}{3} = \tan\dfrac{\pi}{3} = \sqrt{3}$ .
Cosine rule, largest angle opposite the longest side. The $11$ cm side faces the largest angle $C$ : $\cos C = \dfrac{7^2 + 8^2 - 11^2}{2 \times 7 \times 8} = \dfrac{49 + 64 - 121}{112} = -\dfrac{8}{112} = -\dfrac{1}{14}$ . So $C \approx 94.1^\circ$ .
Arc and sector formulas in radians. Arc $= 6 \times 1.4 = 8.4$ cm. Area $= \tfrac{1}{2} \times 36 \times 1.4 = 25.2$ cm $^2$ .
Prove one side, using a common denominator. Start on the left and multiply top and bottom by $1 + \sin\theta$ : $\dfrac{\cos\theta(1 + \sin\theta)}{(1 - \sin\theta)(1 + \sin\theta)} = \dfrac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta} = \dfrac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \dfrac{1 + \sin\theta}{\cos\theta}$ .
Single-wave form. $R = \sqrt{8^2 + 6^2} = 10$ and $\tan\alpha = \dfrac{-6}{8}$ , so $\alpha \approx -0.644$ . The wave is $10\cos(\theta + 0.644)$ and the greatest value is $10$ .
Solve for the principal value, then the second angle. $\sin\theta = 0.75$ , so the principal value is $0.848$ radians. The second solution is $\pi - 0.848 = 2.294$ radians.
Binomial expansion. $(2x - 3)^3 = 8x^3 - 3 \times 4x^2 \times 3 + 3 \times 2x \times 9 - 27 = 8x^3 - 36x^2 + 54x - 27$ .
Perpendicular gradient, then a point. The gradient of $y = 2x + 5$ is $2$ , so the perpendicular gradient is $-\dfrac{1}{2}$ . Then $y + 1 = -\dfrac{1}{2}(x - 3)$ , which is $x + 2y - 1 = 0$ .
Deduce with a general letter. Let the numbers be $2m + 1$ and $2n + 1$ . Their sum is $2m + 2n + 2 = 2(m + n + 1)$ , which has a factor $2$ , so the sum is even.
Factorise the quadratic. $x^2 - 4x + 3 = (x - 1)(x - 3)$ , so the curve crosses the $x$ -axis at $(1, 0)$ and $(3, 0)$ .
Read the wave. Amplitude $= 6$ . Period $= \dfrac{2\pi}{3}$ . The middle line is $y = 1$ , so the range is $-5 \le y \le 7$ .
Resolve with cosine and sine. Horizontal $= 30\cos 55^\circ \approx 17.2$ N. Vertical $= 30\sin 55^\circ \approx 24.6$ N.

What you can now do

use the sine rule, the cosine rule and the area formula on any triangle;
measure angles in radians, and find arc lengths and sector areas;
sketch and read the sine, cosine and tangent graphs, with their periods;
state the exact values of the trigonometric functions at the special angles;
use small-angle approximations and say when they are accurate enough;
use secant, cosecant and cotangent, and the inverse functions;
prove identities with the basic, addition and double-angle formulae;
write a sum of a cosine and a sine as a single wave;
solve trigonometric equations in an interval without losing solutions;
resolve vectors and forces, and find a resultant with the sine and cosine rules.

Your relevance reflection

Return to the prompt in the chapter opener: something that repeats or turns in your own life. Write down the repeating quantity again. Now add three things you could not have said before this chapter: how to write its largest and smallest values, how to model it with a single wave, and how to find when it reaches a chosen level. Finish with one sentence about a quantity you would measure from now on to check a repeating pattern.