Introduction to trigonometry

$\sin 30° = \dfrac{5}{10} = 0.5$ .
$\tan 45° = \dfrac{7}{7} = 1$ .
$O = 20 \sin 40° \approx 20 \times 0.6428 = 12.86$ cm.
$A = 18 \cos 55° \approx 18 \times 0.5736 = 10.32$ cm.
$H = \dfrac{10}{\sin 33°} = \dfrac{10}{0.5446} \approx 18.36$ cm.
$\theta = \sin^{-1}\!\left(\dfrac{6}{13}\right) \approx 27.5°$ .
$\theta = \cos^{-1}\!\left(\dfrac{9}{15}\right) = \cos^{-1}(0.6) \approx 53.1°$ .
$\theta = \tan^{-1}\!\left(\dfrac{12}{5}\right) = \tan^{-1}(2.4) \approx 67.4°$ .
Height $= 50 \sin 60° = 50 \times 0.8660 \approx 43.3$ m.
Height up wall $= 3 \sin 72° \approx 3 \times 0.9511 = 2.85$ m.

$\approx 32.0$ m. Method: the angle of depression equals the angle at the boat. $\tan 38° = \dfrac{25}{d}$ ; $d = \dfrac{25}{\tan 38°} = \dfrac{25}{0.7813} \approx 32.0$ .
$\approx 7.1°$ . Method: $\tan \theta = \dfrac{1}{8} = 0.125$ ; $\theta = \tan^{-1}(0.125)$ .
$\approx 33.7°$ . Method: $\tan \theta = \dfrac{1.2}{1.8} = 0.6\overline{6}$ ; $\theta = \tan^{-1}(0.667)$ .
$\approx 53.1°$ . Method: half-base $= 6$ cm; $\cos \theta = \dfrac{6}{10} = 0.6$ ; $\theta = \cos^{-1}(0.6)$ .
Bearing $\approx 031°$ . Method: $\tan \alpha = \dfrac{9}{15} = 0.6$ ; $\alpha = \tan^{-1}(0.6) \approx 31.0°$ . Bearing from east is $90° - 31° = 59°$ … Correction: bearing is measured clockwise from north. The ship is east then north, so the angle from north $= 90° - \tan^{-1}(\tfrac{9}{15})$ . Actually: from start, east $= 15$ , north $= 9$ . Bearing $= \tan^{-1}(\tfrac{15}{9}) = \tan^{-1}(1.667) \approx 59.0°$ . Bearing $\approx 059°$ .
$41.0$ m. Method: let extra height above the shorter building $= x$ . $\tan 35° = \dfrac{x}{30}$ ; $x = 30 \tan 35° \approx 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

$\sin \theta = \dfrac{O}{H}$ . In a right-angled triangle, the opposite side is always shorter than the hypotenuse (the hypotenuse is the longest side). Therefore $O < H$ , so $\dfrac{O}{H} < 1$ , meaning $\sin \theta < 1$ for acute angles.
$\dfrac{\sin \theta}{\cos \theta} = \dfrac{O/H}{A/H} = \dfrac{O}{H} \times \dfrac{H}{A} = \dfrac{O}{A} = \tan \theta$ .
$\approx 59.3$ m. Method: the river width $w$ is opposite the $56°$ angle; the $40$ m walk is adjacent. $\tan 56° = \dfrac{w}{40}$ ; $w = 40 \tan 56° \approx 59.3$ m.
In a 30-60-90 triangle, the side opposite $30°$ is half the hypotenuse, and the side adjacent to $30°$ is $\dfrac{\sqrt{3}}{2}$ times the hypotenuse. Now $\sin 30° = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . For $60°$ , the opposite and adjacent sides swap: $\cos 60° = \dfrac{\text{adj}_{60}}{H} = \dfrac{\text{opp}_{30}}{H} = \dfrac{1}{2}$ . So $\sin 30° = \cos 60°$ .
(a) $\sin \theta = \dfrac{80}{200} = 0.4$ ; $\theta = \sin^{-1}(0.4) \approx 23.6°$ . (b) Horizontal distance $= \sqrt{200^2 - 80^2} = \sqrt{33600} \approx 183.3$ m. Or: $200 \cos 23.6° \approx 183.3$ m.

$\approx 24.9$ m. Method: let $h$ = height, $d$ = original distance. From the first position: $\tan 28° = \dfrac{h}{d}$ , so $d = \dfrac{h}{\tan 28°}$ . From the closer position: $\tan 42° = \dfrac{h}{d - 20}$ , so $d - 20 = \dfrac{h}{\tan 42°}$ . Substituting: $\dfrac{h}{\tan 28°} - 20 = \dfrac{h}{\tan 42°}$ ; $h\!\left(\dfrac{1}{\tan 28°} - \dfrac{1}{\tan 42°}\right) = 20$ ; $h\!\left(\dfrac{1}{0.5317} - \dfrac{1}{0.9004}\right) = 20$ ; $h(1.8807 - 1.1106) = 20$ ; $h \times 0.7701 = 20$ ; $h \approx 26.0$ m. (Accept $25-26$ m depending on rounding.)
(a) In a regular hexagon, the distance from centre to vertex equals the side length, so $8$ cm. (b) The hexagon splits into $6$ equilateral triangles of side $8$ cm. Area of each $= \dfrac{\sqrt{3}}{4}(8^2) = 16\sqrt{3}$ . Total $= 96\sqrt{3} \approx 166.3$ cm $^2$ .
$\sin^2\theta + \cos^2\theta = \left(\dfrac{O}{H}\right)^2 + \left(\dfrac{A}{H}\right)^2 = \dfrac{O^2 + A^2}{H^2}$ . By Pythagoras, $O^2 + A^2 = H^2$ . Therefore $\dfrac{H^2}{H^2} = 1$ .
Distance travelled in $2$ min $= 250 \times \dfrac{2}{60} = 8.33$ km. (a) Altitude $= 8.33 \sin 15° \approx 8.33 \times 0.2588 \approx 2.16$ km. (b) Horizontal distance $= 8.33 \cos 15° \approx 8.33 \times 0.9659 \approx 8.05$ km.

Tier 1