Trigonometry

From the 30-60-90 triangle, $\sin 30^\circ = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{1}{2}$. ($\dfrac{\sqrt{3}}{2}$ is $\cos 30^\circ$.)

In a 45-45-90 triangle the legs are equal, so $\cos 45^\circ = \dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$ (rationalised). Note $\sin 45^\circ$ has the same value.

The sides of a 30-60-90 triangle are in the ratio $1 : \sqrt{3} : 2$ (opposite 30° : opposite 60° : hypotenuse). With hypotenuse 2, the side opposite $60^\circ$ is $\sqrt{3}$. (1 is the side opposite $30^\circ$.)

In any triangle the longest side is opposite the largest angle. The right angle ($90^\circ$) is the largest angle in a right-angled triangle, so the side opposite it (the hypotenuse) is the longest.

This is the Pythagorean identity: $\sin^2\theta + \cos^2\theta = 1$ for every angle. It follows from Pythagoras applied to a right-angled triangle with hypotenuse 1.

Have the hypotenuse, want the opposite, so use SOH: $\sin 30^\circ = \dfrac{\text{opp}}{\text{hyp}}$, so $\text{opp} = 20 \times \sin 30^\circ = 20 \times 0.5 = 10$ cm. (17.32 cm used $\cos$ (the adjacent); 0.5 cm gave $\sin 30^\circ$ without multiplying.)

The rise (1 m) is opposite the angle and the horizontal run (8 m) is adjacent, so use TOA: $\tan\theta = \dfrac{1}{8} = 0.125$, then $\theta = \tan^{-1}(0.125) \approx 7^\circ$. ($83^\circ$ is the angle at the top of the ramp.)

Have the opposite, want the hypotenuse, so use SOH: $\sin 30^\circ = \dfrac{5}{\text{hyp}}$, so $\text{hyp} = \dfrac{5}{\sin 30^\circ} = \dfrac{5}{0.5} = 10$ cm. Key point: to find the hypotenuse you divide by the ratio. (2.5 cm multiplied $5 \times 0.5$ instead of dividing.)

The flight path (2000 m) is the hypotenuse and the height is opposite the $15^\circ$ angle, so use SOH: $\text{height} = 2000 \times \sin 15^\circ \approx 2000 \times 0.259 \approx 518$ m. ($1932$ m used $\cos$ (the horizontal distance).)

The height splits the triangle into two right-angled triangles, each with the height (8 cm) opposite the base angle and half the base (6 cm) adjacent. Use TOA: $\tan\theta = \dfrac{8}{6} \approx 1.333$, so $\theta = \tan^{-1}(1.333) \approx 53^\circ$. ($37^\circ$ is half of the apex angle.)

Have opposite and hypotenuse, so use SOH: $\sin\theta = \dfrac{5}{10} = 0.5$, then $\theta = \sin^{-1}(0.5) = 30^\circ$. (60° used $\cos^{-1}$ (the complementary angle); 0.5 stopped before taking the inverse.)

Have adjacent and hypotenuse, so use CAH: $\cos\theta = \dfrac{6}{12} = 0.5$, so $\theta = \cos^{-1}(0.5) = 60^\circ$. (30° used $\sin^{-1}$; 0.5 stopped before the inverse.)

The string (60 m) is the hypotenuse and the height is opposite the $50^\circ$ angle, so use SOH: $\text{height} = 60 \times \sin 50^\circ \approx 60 \times 0.766 \approx 46$ m. ($39$ m used $\cos$ (the horizontal distance).)

Opposite and adjacent are linked by tangent ($\tan\theta = \dfrac{\text{opp}}{\text{adj}}$). (Sine and cosine both need the hypotenuse, which is not given.)

Apply the inverse sine: $\theta = \sin^{-1}(0.8) \approx 53.13^\circ \approx 53^\circ$. (37° used $\cos^{-1}(0.8)$ (the complement); 0.8° treated the ratio as the angle, the most common error.)

The 30 m horizontal distance is adjacent and the height is opposite, so use TOA: $\tan 60^\circ = \dfrac{h}{30}$, so $h = 30 \times \tan 60^\circ \approx 30 \times 1.732 \approx 51.96$ m. (25.98 m used $\sin$; 15 m used $\cos$.)

By alternate angles the elevation from the boat is also $30^\circ$. The cliff height (100 m) is opposite and the horizontal distance is adjacent, so $\tan 30^\circ = \dfrac{100}{d}$, giving $d = \dfrac{100}{\tan 30^\circ} \approx 173.21$ m. (57.74 m multiplied $100 \times \tan 30^\circ$ instead of dividing.)

The pole (6 m) is opposite the elevation angle and the shadow (8 m) is adjacent, so use TOA: $\tan\theta = \dfrac{6}{8} = 0.75$, then $\theta = \tan^{-1}(0.75) \approx 37^\circ$. (53° used $\tan^{-1}\left(\dfrac{8}{6}\right)$ (the complementary angle).)

The 50 km path is the hypotenuse, and because the angle is measured from north, the north component is adjacent: $\text{north} = 50 \times \cos 60^\circ = 50 \times 0.5 = 25$ km. (43.3 km used $\sin$ (the east component); 100 km divided instead of multiplying.)

The ladder (5 m) is the hypotenuse and the 3 m distance from the wall is adjacent to the ground angle, so use CAH: $\cos\theta = \dfrac{3}{5} = 0.6$, then $\theta = \cos^{-1}(0.6) \approx 53^\circ$. (37° used $\sin^{-1}\left(\dfrac{3}{5}\right)$ (the angle with the wall).)

The smaller angle is opposite the shorter side (5 cm): $\sin\theta = \dfrac{5}{13} \approx 0.385$, so $\theta = \sin^{-1}(0.385) \approx 23^\circ$. (67° gave the larger acute angle; the other leg is $\sqrt{13^2 - 5^2} = 12$.)

With $\sin\theta = \dfrac{3}{5}$, the opposite is 3 and the hypotenuse is 5. By Pythagoras the adjacent is $\sqrt{5^2 - 3^2} = 4$, so $\cos\theta = \dfrac{4}{5}$ (the classic 3-4-5 triangle). ($\dfrac{3}{4}$ is $\tan\theta$; $\dfrac{5}{4}$ is a reciprocal.)

By alternate angles the elevation from the swimmer is $45^\circ$. With $\tan 45^\circ = \dfrac{100}{d}$ and $\tan 45^\circ = 1$, we get $d = 100$ m. At $45^\circ$ the opposite and adjacent are equal. (141.42 m used $\dfrac{100}{\sin 45^\circ}$, the line-of-sight distance.)

Let $h$ be the height and $d$ the distance from $Q$ to the tower. From $Q$: $h = d\tan 60^\circ = d\sqrt{3}$. From $P$: $h = (d + 50)\tan 30^\circ = \dfrac{d + 50}{\sqrt{3}}$. Setting these equal gives $3d = d + 50$, so $d = 25$ and $h = 25\sqrt{3} \approx 43.3$ m. (25 m gave the distance $d$; 50 m is the distance between $P$ and $Q$.)

$\sin\theta = \dfrac{\text{opp}}{10} = \dfrac{3}{5}$, so the opposite is 6, and by Pythagoras the adjacent is $\sqrt{10^2 - 6^2} = 8$ (a 6-8-10 triangle). Area $= \dfrac{1}{2} \times 6 \times 8 = 24$ cm². (30 cm² used the hypotenuse; 48 cm² forgot to halve.)