Pythagoras' theorem and trigonometry in 3D

$d = \sqrt{36 + 64 + 100} = \sqrt{200} \approx 14.14$ cm.
$d = \sqrt{25 + 25 + 25} = \sqrt{75} = 5\sqrt{3} \approx 8.66$ cm.
$\tan 40° = \frac{25}{d}$ , so $d = \frac{25}{\tan 40°} \approx \frac{25}{0.8391} \approx 29.79$ m.
$\tan 50° = \frac{80}{d}$ , so $d = \frac{80}{\tan 50°} \approx \frac{80}{1.1918} \approx 67.13$ m.
North $= 10\cos 45° \approx 7.07$ km; East $= 10\sin 45° \approx 7.07$ km.
Percentage error $= \frac{0.2}{8.0} \times 100 = 2.5\%$ . Squared value error $\approx 2 \times 2.5\% = 5\%$ .
$d = \sqrt{144 + 25 + 16} = \sqrt{185} \approx 13.60$ cm.
Half-diagonal of base $= \frac{10\sqrt{2}}{2} = 5\sqrt{2} \approx 7.071$ cm. Slant edge $= \sqrt{12^2 + (5\sqrt{2})^2} = \sqrt{144 + 50} = \sqrt{194} \approx 13.93$ cm.

Unfold the $6 \times 3$ wall and the $4 \times 3$ wall into a single rectangle $10 \times 3$ . The spider walks diagonally: $\sqrt{10^2 + 3^2} = \sqrt{109} \approx 10.44$ m. (Other unfoldings give longer paths; the shortest is $\sqrt{(6+3)^2 + 4^2} = \sqrt{97} \approx 9.85$ m by unfolding the $6 \times 3$ wall with the floor. Check all configurations; the minimum is approximately $9.85$ m.)
Leg 1: N $= 5\cos 30° \approx 4.33$ km, E $= 5\sin 30° = 2.5$ km. Leg 2: N $= -8\cos 60° = -4$ km (south), E $= 8\sin 60° \approx 6.93$ km. Total: N $\approx 0.33$ km, E $\approx 9.43$ km. Distance $= \sqrt{0.33^2 + 9.43^2} \approx 9.44$ km. Bearing $= \tan^{-1}(9.43/0.33) \approx 88°$ .
Boat 1: $d_1 = \frac{60}{\tan 45°} = 60$ m from cliff base. Boat 2: $d_2 = \frac{60}{\tan 30°} \approx 103.92$ m. Distance between boats $\approx 103.92 - 60 = 43.92$ m.
Let height $= h$ and distance from A to the base $= x$ . From A: $\tan 22° = h/x$ . From B: $\tan 15° = h/(x + 500)$ . So $x = h/\tan 22°$ and $x + 500 = h/\tan 15°$ . Subtracting: $500 = h(1/\tan 15° - 1/\tan 22°) = h(3.732 - 2.475) = 1.257h$ . So $h \approx 397.8$ m.
$V = \pi(100)(20) = 2000\pi \approx 6283.19$ cm $^3$ . Error in $r$ : $\frac{0.3}{10} = 3\%$ , doubled for $r^2$ : $6\%$ . Error in $h$ : $\frac{0.5}{20} = 2.5\%$ . Total $\approx 8.5\%$ .
(a) Rope $= \sqrt{2.5^2 + 1.5^2} = \sqrt{8.5} \approx 2.92$ m. (b) $\tan\theta = \frac{2.5}{1.5}$ , so $\theta = \tan^{-1}(1.667) \approx 59.0°$ .

First, in the base rectangle, we form a right triangle with legs $l$ and $w$ to get the base diagonal $d_b = \sqrt{l^2 + w^2}$ . Second, this base diagonal becomes one leg of a new right triangle whose other leg is the height $h$ and whose hypotenuse is the space diagonal. Applying Pythagoras again: $d = \sqrt{d_b^2 + h^2} = \sqrt{l^2 + w^2 + h^2}$ .
Leg 1 (A to B): N $= 200\cos 70° \approx 68.40$ km, E $= 200\sin 70° \approx 187.94$ km. Leg 2 (B to C): N $= -150\cos 20° \approx -140.95$ km, E $= 150\sin 20° \approx 51.30$ km. C relative to A: N $\approx -72.55$ km, E $\approx 239.24$ km. Return C to A: N $\approx 72.55$ , W $\approx 239.24$ . Distance $= \sqrt{72.55^2 + 239.24^2} \approx 250.0$ km. Bearing from C to A: $\theta = \tan^{-1}(239.24/72.55) \approx 73.1°$ west of north $= 360° - 73.1° = 286.9°$ .
Building height $h$ . When shadow $= 30$ : $\tan 55° = h/30$ , so $h = 30\tan 55° \approx 42.84$ m. When shadow $= 45$ : $\tan\alpha = 42.84/45$ , so $\alpha = \tan^{-1}(0.952) \approx 43.6°$ . The angle decreased from $55°$ to $43.6°$ , so the sun fell (moved lower in the sky).
For $V = \frac{4}{3}\pi r^3$ , the radius is cubed, so the percentage error in $V$ is approximately $3$ times the percentage error in $r$ (not double). For example, a $1\%$ error in $r$ gives roughly a $3\%$ error in volume. The multiplier equals the exponent of the variable in the formula.

Base half-diagonal $= \frac{12\sqrt{2}}{2} = 6\sqrt{2}$ . Slant height $= 10$ (distance from midpoint of base edge to apex). Half base edge $= 6$ . Vertical height from base edge midpoint: $h_{\text{face}} = \sqrt{10^2 - 6^2} = 8$ cm. (a) Vertical height of pyramid: the midpoint of a base edge is $6$ cm from the centre (for a $12 \times 12$ base). So $H = \sqrt{10^2 - 6^2} = 8$ cm. (Alternatively, using the half-diagonal: $H = \sqrt{10^2 - (6\sqrt{2})^2}$ … but slant height goes from base edge midpoint, not corner. Let’s recalculate. Slant height $l = 10$ is from midpoint of a base edge to apex. Half base $= 6$ . Distance from centre to midpoint of edge $= 6$ . $H = \sqrt{l^2 - 6^2} = \sqrt{100 - 36} = 8$ cm.) (b) Angle between slant face and base: $\tan\alpha = H/6 = 8/6$ , so $\alpha = \tan^{-1}(4/3) \approx 53.1°$ . (c) Slant edge (corner to apex): distance from centre to corner $= 6\sqrt{2}$ . Slant edge $= \sqrt{8^2 + (6\sqrt{2})^2} = \sqrt{64 + 72} = \sqrt{136} \approx 11.66$ cm. Angle with base: $\tan\beta = 8/(6\sqrt{2}) = 8/8.485$ , so $\beta \approx 43.3°$ .
Let A be at the origin, B at $(300, 0)$ . Drone bearing $060°$ from A means the drone’s horizontal position is along direction $060°$ from A. Let horizontal distance from A to point below drone $= d_A$ . Then $d_A \tan 40°...$ — actually the drone is at some point $(x, y, h)$ . From A: bearing $060°$ means $x = d_A \sin 60°$ , $y = d_A \cos 60°$ and $\tan 40° = h/d_A$ . From B at $(300, 0)$ : $\tan 55° = h/d_B$ where $d_B$ is horizontal distance from B to drone. Using $d_A = h/\tan 40°$ and $d_B = h/\tan 55°$ , and the geometry: $x = d_A \sin 60° = h\sin 60°/\tan 40°$ , $y = d_A \cos 60° = h\cos 60°/\tan 40°$ . Also $d_B^2 = (x - 300)^2 + y^2 = h^2/\tan^2 55°$ . Substituting and solving numerically gives $h \approx 218$ m. (Accept reasonable numerical solutions with clear working.)
Cube side $= 2r$ (the sphere diameter equals the cube side). Space diagonal of cube $= 2r\sqrt{3}$ . Diameter of sphere $= 2r$ . Ratio $= \frac{2r\sqrt{3}}{2r} = \sqrt{3}:1$ .
Nominal area $= \frac{1}{2}(50)(80)\sin 60° = 2000 \times 0.8660 \approx 1732$ m $^2$ . Length errors: $\frac{0.5}{50} = 1\%$ and $\frac{0.5}{80} = 0.625\%$ . Since each length appears to the first power, their contributions are $1\%$ and $0.625\%$ . Angle error: $1°$ in $60°$ . The sensitivity factor is $\frac{d}{dC}(\sin C) \cdot \frac{C}{\sin C}$ . At $C = 60°$ : $\frac{\cos 60°}{\sin 60°} \times 1° \times \frac{\pi}{180} \approx 0.577 \times 0.01745 \approx 1.007\%$ . Total error $\approx 1 + 0.625 + 1.007 \approx 2.6\%$ , i.e. $\pm 45$ m $^2$ . The angle error contributes roughly as much as the length errors combined.

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