Exam practice

Opposite $\varphi$ (at C) is AB. Perpendicular p = AB, base b = BC, hypotenuse h = AC.

$\sin\varphi = \dfrac{AB}{AC}$, $\cos\varphi = \dfrac{BC}{AC}$, $\tan\varphi = \dfrac{AB}{BC}$, $\operatorname{cosec}\varphi = \dfrac{AC}{AB}$, $\sec\varphi = \dfrac{AC}{BC}$, $\cot\varphi = \dfrac{BC}{AB}$.

Opposite β (at A) is BC. Perpendicular p = BC, base b = AB, hypotenuse h = AC.

$\sin\beta = \dfrac{BC}{AC}$, $\cos\beta = \dfrac{AB}{AC}$, $\tan\beta = \dfrac{BC}{AB}$, $\operatorname{cosec}\beta = \dfrac{AC}{BC}$, $\sec\beta = \dfrac{AC}{AB}$, $\cot\beta = \dfrac{AB}{BC}$.
Notice that $\sin\beta = \cos\varphi$ and $\cos\beta = \sin\varphi$: the two acute angles share their sides with the roles swapped.

First the hypotenuse: $PR = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$. From β at P: p = QR = 8, b = PQ = 6.
$\sin\beta = \dfrac{8}{10} = \dfrac45$, $\cos\beta = \dfrac{6}{10} = \dfrac35$, $\tan\beta = \dfrac86 = \dfrac43$.

$\operatorname{cosec}\beta = \dfrac54$, $\sec\beta = \dfrac53$, $\cot\beta = \dfrac34$.

From α at R: p = PQ = 6, b = QR = 8, h = 10.
$\sin\alpha = \dfrac35$, $\cos\alpha = \dfrac45$, $\tan\alpha = \dfrac34$, $\operatorname{cosec}\alpha = \dfrac53$, $\sec\alpha = \dfrac54$, $\cot\alpha = \dfrac43$.

$\dfrac{p}{20} = \dfrac{6}{10}$, so p = 12. Then $b = \sqrt{20^2 - 12^2} = \sqrt{400 - 144} = \sqrt{256} =$ 16.

• i.With p = 5 and b = 12 (or 5k and 12k): $h = \sqrt{25 + 144} = \sqrt{169} =$ 13 (or 13k).
• ii.$\sin\theta = \dfrac{5}{13}$, $\cos\theta = \dfrac{12}{13}$.

With b = 8k and h = 17k: $p = \sqrt{(17k)^2 - (8k)^2} = \sqrt{225k^2} = 15k$.
$\sin\theta = \dfrac{15}{17}$, $\tan\theta = \dfrac{15}{8}$. Check: $\left(\dfrac{15}{17}\right)^2 + \left(\dfrac{8}{17}\right)^2 = \dfrac{225 + 64}{289} = 1$ ✓.

$\dfrac{p^2}{h^2} + \dfrac{b^2}{h^2} = \dfrac{p^2 + b^2}{h^2} = \dfrac{h^2}{h^2} =$ 1.

$1 - \dfrac{b^2}{h^2} = \dfrac{h^2 - b^2}{h^2} = \dfrac{p^2}{h^2} =$ $\sin^2 A$.

$\dfrac{p}{h} \div \dfrac{b}{h} = \dfrac{p}{h} \times \dfrac{h}{b} = \dfrac{p}{b} =$ tan θ.

$\dfrac{p^2}{b^2} + 1 = \dfrac{p^2 + b^2}{b^2} = \dfrac{h^2}{b^2} =$ $\sec^2\theta$.

$\dfrac{b}{h} \div \dfrac{p}{h} = \dfrac{b}{h} \times \dfrac{h}{p} = \dfrac{b}{p} =$ $\cot\theta$.

$\dfrac{h^2}{p^2} - \dfrac{b^2}{p^2} = \dfrac{h^2 - b^2}{p^2} = \dfrac{p^2}{p^2} =$ 1.

LHS $= \dfrac{p^2}{h^2} + \dfrac{b^2}{h^2} = \dfrac{p^2 + b^2}{h^2} = \dfrac{h^2}{h^2} = 1 =$ RHS, since $p^2 + b^2 = h^2$ (Pythagoras). Proved.

LHS $= 1 - \dfrac{p^2}{h^2} = \dfrac{h^2 - p^2}{h^2} = \dfrac{b^2}{h^2} = \cos^2\theta =$ RHS. Proved.

LHS $= \dfrac{p/h}{b/h} = \dfrac{p}{h} \times \dfrac{h}{b} = \dfrac{p}{b} = \tan\theta =$ RHS. Proved.

LHS $= \dfrac{h^2}{b^2} - 1 = \dfrac{h^2 - b^2}{b^2} = \dfrac{p^2}{b^2} = \tan^2\theta =$ RHS. Proved.

LHS $= \dfrac{h^2}{p^2} - \dfrac{b^2}{p^2} = \dfrac{h^2 - b^2}{p^2} = \dfrac{p^2}{p^2} = 1 =$ RHS. Proved.

The plane is the hypotenuse (29 ft) and the building’s height the perpendicular (20 ft). Distance $= \sqrt{29^2 - 20^2} = \sqrt{841 - 400} = \sqrt{441} =$ 21 ft.

$\tan\theta = \dfrac{20}{21}$, $\sin\theta = \dfrac{20}{29}$, $\cos\theta = \dfrac{21}{29}$. The plane rises a little less than 1 ft for every foot across: θ is a bit under 45∘°.

$AB = \sqrt{40^2 - 24^2} = \sqrt{1600 - 576} = \sqrt{1024} =$ 32 m.

$\tan\theta = \dfrac{32}{24} = \dfrac43$, $\sin\theta = \dfrac{32}{40} = \dfrac45$, $\cos\theta = \dfrac{24}{40} = \dfrac35$. The 3–4–5 triangle again, eight times bigger.

1

$\dfrac12$

$\dfrac{1}{\sqrt3}$ (about 0.577)

$\dfrac{1}{\tan 45^\circ} =$ 1

$\dfrac{1}{\sin 45^\circ} =$ $\sqrt2$

$\tan\theta = \dfrac{4}{4\sqrt3} = \dfrac{1}{\sqrt3} = \tan 30^\circ$, so $\theta = 30^\circ$.

$\sin\theta = \dfrac{7\sqrt3}{14} = \dfrac{\sqrt3}{2} = \sin 60^\circ$, so $\theta = 60^\circ$.

$h = \sqrt{50} = 5\sqrt2$ cm, so $\cos\theta = \dfrac{5}{5\sqrt2} = \dfrac{1}{\sqrt2} = \cos 45^\circ$: $\theta = 45^\circ$.

$\tan\theta = \dfrac{x}{x} = 1 = \tan 45^\circ$, so $\theta = 45^\circ$.

$p = 10 \sin 60^\circ = 10 \times \dfrac{\sqrt3}{2} =$ $5\sqrt3$ cm (about 8.66 cm); $b = 10 \cos 60^\circ = 10 \times \dfrac12 =$ 5 cm.

$\tan 45^\circ = \dfrac{15}{b}$ gives b = 15 ft; $\sin 45^\circ = \dfrac{15}{h}$ gives $h = 15\sqrt2$ ft (about 21.21 ft).

$p = 36 \tan 30^\circ = \dfrac{36}{\sqrt3} =$ $12\sqrt3$ inches (about 20.78 in); $h = \dfrac{36}{\cos 30^\circ} = \dfrac{72}{\sqrt3} =$ $24\sqrt3$ inches (about 41.57 in). Check: h = 2p, as it must be opposite 30∘°.

$\dfrac12 + \dfrac12 =$ 1

$\dfrac{1}{\sqrt3} + \dfrac{\sqrt3}{2} = \dfrac{2 + 3}{2\sqrt3} =$ $\dfrac{5}{2\sqrt3}$ $\left(= \dfrac{5\sqrt3}{6} \approx 1.443\right)$

$\dfrac12 + \dfrac{\sqrt3}{2} \times \dfrac{1}{\sqrt3} = \dfrac12 + \dfrac12 =$ 1 (multiply before adding).

$\dfrac{\frac{\sqrt3}{2} + \frac{\sqrt3}{2}}{\sqrt3} = \dfrac{\sqrt3}{\sqrt3} =$ 1

LHS $= \left(\dfrac{1}{\sqrt2}\right)^2 + \left(\dfrac{1}{\sqrt2}\right)^2 = \dfrac12 + \dfrac12 = 1 =$ RHS. Proved.

$$\begin{aligned} \text{LHS} &= \frac{2 \times \frac{1}{\sqrt3}}{1 - \frac13} = \frac{2}{\sqrt3} \div \frac23 \\ &= \frac{2}{\sqrt3} \times \frac32 = \frac{3}{\sqrt3} = \sqrt3 \end{aligned}$$
$= \tan 60^\circ =$ RHS. Proved.

$$\begin{aligned} \text{LHS} &= \frac{\sqrt3 - \frac{1}{\sqrt3}}{1 + \sqrt3 \times \frac{1}{\sqrt3}} = \frac{\frac{3 - 1}{\sqrt3}}{2} \\ &= \frac{2}{\sqrt3} \times \frac12 = \frac{1}{\sqrt3} \end{aligned}$$
$= \tan 30^\circ =$ RHS. Proved.

LHS = 1 − 1 = 0. RHS $= 1 - 2 \times \left(\dfrac{1}{\sqrt2}\right)^2 = 1 - 2 \times \dfrac12 = 0$. LHS = RHS. Proved.

With the ground, at F: $\sin\theta = \dfrac{30\sqrt3}{60} = \dfrac{\sqrt3}{2} = \sin 60^\circ$, so the angle is 60∘°. The angles of △FWT add up to 180∘°, so the angle at the wall is $180^\circ - 90^\circ - 60^\circ = 30^\circ$.
60∘° with the ground and 30∘° with the wall.

$\tan\theta = \dfrac{36}{12\sqrt3} = \dfrac{3}{\sqrt3} = \sqrt3 = \tan 60^\circ$, so $\theta = 60^\circ$. The tower is taller than the distance, so the angle had to be more than 45∘°. (The other angle of the triangle, at the top, between the tower and the line of sight, is 30∘°.)

• i.Both acute angles 45∘° makes the triangle isosceles: $\tan 45^\circ = \dfrac{x}{40.5} = 1$, so x = 40.5: the tree is 40.5 ft tall.
• ii.$VJ = \dfrac{40.5}{\cos 45^\circ} = 40.5\sqrt2 \approx$ 57.28 ft.

$\sin 60^\circ = \dfrac{12}{l}$, so $l = \dfrac{12}{\sqrt3/2} = \dfrac{24}{\sqrt3} = 8\sqrt3 \approx$ 13.86 ft ($8\sqrt3$ ft).

Perpendicular p = AC (opposite B), base b = AB, hypotenuse h = BC.

$\sin B = \dfrac{AC}{BC}$, $\cos B = \dfrac{AB}{BC}$, $\tan B = \dfrac{AC}{AB}$, $\operatorname{cosec} B = \dfrac{BC}{AC}$, $\sec B = \dfrac{BC}{AB}$, $\cot B = \dfrac{AB}{AC}$.

$p = \sqrt{20^2 - 12^2} = \sqrt{256} = 16$, so $\tan B = \dfrac{16}{12} = \dfrac43$.

$\sin B = \dfrac{16}{20} = \dfrac45$, $\cos B = \dfrac{12}{20} = \dfrac35$. So $\sin^2 B = \dfrac{16}{25}$, $\cos^2 B = \dfrac{9}{25}$, $\tan^2 B = \dfrac{16}{9}$. Check: $\dfrac{16}{25} + \dfrac{9}{25} = 1$ ✓.

From 60∘° at X: p = YZ = 8 cm.
$\tan 60^\circ = \dfrac{8}{XY}$, so $XY = \dfrac{8}{\sqrt3} = \dfrac{8\sqrt3}{3} \approx$ 4.62 cm.
$\sin 60^\circ = \dfrac{8}{ZX}$, so $ZX = \dfrac{16}{\sqrt3} = \dfrac{16\sqrt3}{3} \approx$ 9.24 cm.

$\sin^2 X + \cos^2 X = \left(\dfrac{\sqrt3}{2}\right)^2 + \left(\dfrac12\right)^2 = \dfrac34 + \dfrac14 = 1$. Proved.

$\angle Z = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. $\sin Z = \dfrac12$, $\cos Z = \dfrac{\sqrt3}{2}$, $\tan Z = \dfrac{1}{\sqrt3}$. (From the sides: $\sin Z = \dfrac{XY}{ZX} = \dfrac{8/\sqrt3}{16/\sqrt3} = \dfrac12$ ✓.)

$\sin^2 Z = \dfrac14$, $\cos^2 Z = \dfrac34$, $\tan^2 Z = \dfrac13$.

$\tan 30^\circ = \dfrac{h}{30}$, so $h = \dfrac{30}{\sqrt3} = 10\sqrt3 \approx$ 17.32 ft ($10\sqrt3$ ft).

At 45∘° the distance equals the height: $10\sqrt3 \approx 17.32$ ft from the foot. So walk $30 - 10\sqrt3 \approx$ 12.68 ft towards it.

$\tan\theta = \dfrac{30\sqrt3}{30} = \sqrt3 = \tan 60^\circ$, so $\theta = 60^\circ$. Tripling the height took the angle from 30∘° to 60∘°, not to 90∘°: the angle doesn’t grow in step with the height. (This bamboo grew $20\sqrt3 \approx 34.6$ ft, about 10.6 m, in 4 days: faster than any real bamboo. The fastest kinds grow up to about 91 cm in a day.)