Trigonometry Applications (Bearings & 3D)

A reverse (back) bearing differs by $180^\circ$. Since $\theta < 180^\circ$, adding $180^\circ$ keeps the result below $360^\circ$, so the bearing of A from B is $\theta + 180^\circ$. (If $\theta$ were larger than $180^\circ$ you would subtract instead.)

The two legs are perpendicular, so the displacement is the hypotenuse: $d = \sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15$ km (a 9-12-15 triangle). (21 km added the legs; 225 km forgot the square root.)

The tower height is the opposite side: $h = 50\tan 35^\circ \approx 50(0.700) \approx 35$ m. (29 m used $50\sin 35^\circ$; 71 m divided by $\tan 35^\circ$ instead of multiplying.)

The horizontal line at the cliff-top and the horizontal at the boat are parallel, so the angle of depression and the angle of elevation are alternate angles and are equal: $25^\circ$. (65° is its complement, not the alternate angle.)

A bearing of $120^\circ$ is $30^\circ$ past due east, toward the south. The southward component is $100\sin(120^\circ - 90^\circ) = 100\sin 30^\circ = 50$ km. (87 km is the eastward component $100\cos 30^\circ$.)

The distance is the hypotenuse and the bearing is measured from north, so east is the opposite side: $\text{east} = 40 \times \sin 60^\circ = 40 \times \dfrac{\sqrt{3}}{2} = 20\sqrt{3}$ km. (20 km is the north component, $40 \times \cos 60^\circ$; 40 km is the hypotenuse.)

The bearing measured from north satisfies $\tan\theta = \dfrac{\text{east}}{\text{north}} = \dfrac{3}{4}$, so $\theta = \tan^{-1}(0.75) \approx 37^\circ$, a true bearing of 037°. (053° used $\dfrac{4}{3}$, the angle from east; 217° added 180°.)

A reverse bearing differs by 180°: 050° + 180° = 230°. (050° is in the NE quadrant and 230° in the opposite SW quadrant. 310° = 360° − 50°, the wrong rule.)

At Q, the direction back to P is the reverse of 070°, which is 250°; the direction to R is 160°. The angle PQR $= 250^\circ - 160^\circ = 90^\circ$ (equivalently, the turn is $160^\circ - 70^\circ = 90^\circ$). (70° and 160° are the leg bearings, not the angle.)

The two legs are perpendicular, so by Pythagoras $d = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50$ km (a 30-40-50 triangle). (70 km added the legs; $\sqrt{70}$ forgot to square.)

By Pythagoras the unknown side is $\sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm (a 5-12-13 triangle). ($\sqrt{194}$ added the squares instead of subtracting.)

$d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ cm. (11 cm added the three dimensions; $\sqrt{13}$ used only two; 49 cm forgot the square root.)

The base half-diagonal is $\dfrac{1}{2} \times 6\sqrt{2} = 3\sqrt{2}$ cm. The slant edge is the hypotenuse of a right triangle with the height and this half-diagonal: $\sqrt{(3\sqrt{2})^2 + 4^2} = \sqrt{18 + 16} = \sqrt{34} \approx 5.83$ cm. (Using half a side (3) instead of half the diagonal gives the slant height, $5$ cm.)

$d = \sqrt{6^2 + 6^2 + 6^2} = \sqrt{108} = 6\sqrt{3}$ cm. ($6\sqrt{2}$ is the face diagonal (two dimensions only); $18$ omitted the square root.)

Let the near point be $x$ m from the base. Then $\tan 60^\circ = \dfrac{h}{x}$ and $\tan 30^\circ = \dfrac{h}{x + 20}$, so $x = \dfrac{h}{\sqrt{3}}$ and $x + 20 = h\sqrt{3}$. Subtracting gives $h\sqrt{3} - \dfrac{h}{\sqrt{3}} = \dfrac{2h}{\sqrt{3}} = 20$, so $h = 10\sqrt{3} \approx 17.3$ m. ($\dfrac{20}{\sqrt{3}}$ solved only one of the triangles.)

The peak is at the centre, so each slope rises above half the span: $\tan\theta = \dfrac{\text{rise}}{\text{half-span}} = \dfrac{3}{4}$, giving $\theta = \tan^{-1}(0.75) \approx 37^\circ$. ($21^\circ$ used the full span $\left(\tfrac{3}{8}\right)$ instead of halving it; $53^\circ$ used $\tfrac{4}{3}$.)

The height is opposite and the 20 m distance is adjacent: $h = 20 \times \tan 60^\circ = 20\sqrt{3}$ m. ($\dfrac{20}{\sqrt{3}}$ used $\dfrac{1}{\tan 60^\circ}$; 20 m used $\tan 45^\circ$.)

The diagonal across the field is $\sqrt{30^2 + 40^2} = 50$ m. Then $h = 50 \times \tan 45^\circ = 50 \times 1 = 50$ m. (30 m and 40 m use a single side instead of the diagonal; 70 m added the sides.)

The base diagonal is $\sqrt{6^2 + 8^2} = 10$ cm. With $\tan 45^\circ = \dfrac{h}{10} = 1$, the height is $h = 10$ cm. ($10\sqrt{2}$ cm is the space diagonal; 14 cm added the base edges.)

By alternate angles the elevation from the worm is also 30°. The height (40 m) is opposite and the horizontal distance is adjacent: $\tan 30^\circ = \dfrac{40}{d}$, so $d = \dfrac{40}{\tan 30^\circ} = 40\sqrt{3}$ m. ($40\sqrt{2}$ m used 45°; 40 m used $\tan 45^\circ$.)

$\tan(\text{bearing}) = \dfrac{\text{east}}{\text{north}} = \dfrac{15}{8}$, so the bearing $= \tan^{-1}(1.875) \approx 62^\circ$, giving 062° (NE quadrant). (028° used $\dfrac{8}{15}$, the angle from east.)

The angle at the lighthouse is $120^\circ - 30^\circ = 90^\circ$, and the two distances are equal (20 km), forming a right-angled isosceles triangle: $ST = \sqrt{20^2 + 20^2} = \sqrt{800} = 20\sqrt{2}$ km. (20 km is one leg; 40 km doubled.)

The base half-diagonal is $\dfrac{1}{2} \times 8\sqrt{2} = 4\sqrt{2} \approx 5.66$ cm. At the base corner the height is opposite and the half-diagonal adjacent: $\tan\theta = \dfrac{6}{4\sqrt{2}} \approx 1.06$, so $\theta \approx 47^\circ$. (Using half a base side instead of half the diagonal gives the wrong angle.)

North is the adjacent side (the bearing is measured from north): $\text{north} = 20 \times \cos 30^\circ = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ km. (10 km is the east component, $20 \times \sin 30^\circ$; 20 km is the hypotenuse.)

Far boat: $d_1 = \dfrac{60}{\tan 30^\circ} = 60\sqrt{3}$ m. Close boat: $d_2 = \dfrac{60}{\tan 60^\circ} = \dfrac{60}{\sqrt{3}} = 20\sqrt{3}$ m. Distance apart $= 60\sqrt{3} - 20\sqrt{3} = 40\sqrt{3}$ m. (40 m dropped the $\sqrt{3}$; 60 m is the cliff height.)