Pythagoras' theorem and trigonometry applications

$c = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
$a = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ m.
$d = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ m.
$\sin 40° \approx 0.64$ , $\cos 40° \approx 0.77$ , $\tan 40° \approx 0.84$ .
$\text{opp} = 20 \sin 50° \approx 20 \times 0.7660 = 15.32$ cm.
$\theta = \tan^{-1}\!\left(\dfrac{3}{4}\right) = \tan^{-1}(0.75) \approx 36.87° \approx 36.9°$ .
$d = \sqrt{(4-1)^2 + (7-3)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ units.
$d = \sqrt{(3-(-2))^2 + (-7-5)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ units.
$\sin\theta = \dfrac{5.5}{6}$ , so $\theta = \sin^{-1}(0.9167) \approx 66.4°$ .
$d = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ m.

$\sin\theta = \dfrac{40}{65} = \dfrac{8}{13}$ , so $\theta = \sin^{-1}(0.6154) \approx 38.0°$ .
$\tan 18° = \dfrac{50}{d}$ , so $d = \dfrac{50}{\tan 18°} \approx \dfrac{50}{0.3249} \approx 153.9$ m.
North: $8 \cos 40° \approx 8 \times 0.7660 = 6.13$ km. East from first leg: $8 \sin 40° \approx 8 \times 0.6428 = 5.14$ km. Total east: $5.14 + 6 = 11.14$ km. Total north: $6.13$ km.
$AB = 6$ , $BC = \sqrt{(6-3)^2 + (0-5)^2} = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ , $AC = \sqrt{9 + 25} = \sqrt{34} \approx 5.83$ . Perimeter $\approx 6 + 5.83 + 5.83 = 17.66$ units.
$\tan\theta = \dfrac{2.4}{1.8} = \dfrac{4}{3}$ , so $\theta = \tan^{-1}(1.333) \approx 53.1°$ .
Space diagonal: $d^2 = l^2 + w^2 + h^2$ . So $15^2 = 5^2 + 12^2 + h^2$ , giving $225 = 25 + 144 + h^2$ , thus $h^2 = 56$ and $h = \sqrt{56} \approx 7.48$ cm.
South: $20 \cos 60° = 20 \times 0.5 = 10$ km (bearing $120°$ is $60°$ from south). East: $20 \sin 60° = 20 \times 0.8660 \approx 17.32$ km.
Let extra height above the shorter building be $x$ . $\tan 35° = \dfrac{x}{30}$ , so $x = 30 \tan 35° \approx 30 \times 0.7002 = 21.0$ m. Total height $= 20 + 21.0 = 41.0$ m.

On the Cartesian plane, the horizontal distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $|x_2 - x_1|$ and the vertical distance is $|y_2 - y_1|$ . These form the two shorter sides of a right-angled triangle. Applying Pythagoras’ theorem: $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ , giving $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .
Observer A: height $= 500 \tan 28° \approx 500 \times 0.5317 = 265.9$ m. Observer B: height $= 700 \tan 20° \approx 700 \times 0.3640 = 254.8$ m. The heights differ ( $265.9 \neq 254.8$ ), so either the measurements are imprecise or the helicopter is not directly above the assumed point.
$\tan 62° = \dfrac{w}{50}$ , where $w$ is the river width. $w = 50 \tan 62° \approx 50 \times 1.8807 = 94.0$ m.
This is the standard triangle inequality. For any triangle with vertices on the coordinate plane, the shortest path between two points is the straight line (the side). Going via a third point is longer, so $AB + BC > AC$ (and cyclic permutations). A rigorous proof uses the Cauchy—Schwarz inequality or the properties of the Euclidean metric.
$d = \sqrt{(-5-3)^2 + (1-7)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ grid units. Actual distance $= 10 \times 2 = 20$ km.

Height: $h = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm. Angle between slant and base: $\cos\alpha = \dfrac{5}{13}$ , so $\alpha = \cos^{-1}(0.3846) \approx 67.4°$ .
The western lighthouse is at $W$ and the eastern at $E$ , 10 km apart. Bearing $320°$ to $W$ means the angle from north is $320°$ , so the line from the ship to $W$ makes $40°$ west of north. Bearing $050°$ to $E$ means $50°$ east of north. The ship is south of the coastline. At the ship, the angle $\angle WSE = 360° - 320° + 50° = 90°$ . Using the right triangle: let perpendicular distance $= h$ . $\tan 40° = \dfrac{d_W}{h}$ and $\tan 50° = \dfrac{d_E}{h}$ where $d_W + d_E = 10$ . So $h(\tan 40° + \tan 50°) = 10$ , giving $h = \dfrac{10}{\tan 40° + \tan 50°} = \dfrac{10}{0.8391 + 1.1918} \approx \dfrac{10}{2.0309} \approx 4.92$ km.
Space diagonal: $d = \sqrt{s^2 + s^2 + s^2} = \sqrt{3s^2} = s\sqrt{3}$ . The base diagonal is $s\sqrt{2}$ . The angle $\alpha$ between the space diagonal and the base satisfies $\tan\alpha = \dfrac{s}{s\sqrt{2}} = \dfrac{1}{\sqrt{2}}$ , so $\alpha = \tan^{-1}\!\left(\dfrac{1}{\sqrt{2}}\right) \approx 35.3°$ .
$PA = \sqrt{5^2 + 4^2} = \sqrt{41} \approx 6.40$ km (within 7 km — yes). $PB = \sqrt{(5-8)^2 + 4^2} = \sqrt{9 + 16} = 5$ km (within 7 km — yes). $PC = \sqrt{(5-3)^2 + (4-6)^2} = \sqrt{4 + 4} = \sqrt{8} \approx 2.83$ km (within 7 km — yes). The phone can receive signal from all three towers.

Tier 1