Right Triangle Trigonometry

1. Right angle at $C$. $AC = 5$ (adjacent to $A$, opposite $B$), $BC = 12$ (opposite $A$, adjacent to $B$), $AB = 13$ (hypotenuse).

(a) Ratios for angle $A$:

$\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{12}{13}$

$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AC}{AB} = \frac{5}{13}$

$\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AC} = \frac{12}{5}$

(b) Ratios for angle $B$:

$\sin B = \frac{AC}{AB} = \frac{5}{13}, \quad \cos B = \frac{BC}{AB} = \frac{12}{13}, \quad \tan B = \frac{AC}{BC} = \frac{5}{12}$

(c) $\sin A = \frac{12}{13} = \cos B$ and $\cos A = \frac{5}{13} = \sin B$.

The sine of one acute angle equals the cosine of the other — this is always true for complementary angles ($A + B = 90°$). This is why "cosine" comes from "complement's sine."

2. (a) Angle $= 35°$, hypotenuse $= 20$, find opposite side $x$:

$\sin 35° = \frac{x}{20} \implies x = 20 \sin 35° \approx 20(0.5736) \approx 11.5$

(b) Angle $= 58°$, adjacent $= 9$, find hypotenuse $h$:

$\cos 58° = \frac{9}{h} \implies h = \frac{9}{\cos 58°} \approx \frac{9}{0.5299} \approx 17.0$

3. (a) Opposite $= 7$, hypotenuse $= 25$:

$\sin \theta = \frac{7}{25} = 0.28 \implies \theta = \sin^{-1}(0.28) \approx 16°$

(b) Adjacent $= 8$, opposite $= 15$:

$\tan \theta = \frac{15}{8} = 1.875 \implies \theta = \tan^{-1}(1.875) \approx 62°$

4. The ladder, wall, and ground form a right triangle. The ladder (hypotenuse) $= 15$ ft, base $= 6$ ft.

(a) Let $\theta$ = angle with the ground. Adjacent side $= 6$, hypotenuse $= 15$:

$\cos \theta = \frac{6}{15} = 0.4 \implies \theta = \cos^{-1}(0.4) \approx 66°$

(b) Height up the wall (opposite side):

$h = \sqrt{15^2 - 6^2} = \sqrt{225 - 36} = \sqrt{189} \approx 13.7 \text{ ft}$

Or using the angle: $h = 15\sin 66° \approx 15(0.9135) \approx 13.7$ ft.

5. (a) Diagram: right triangle with the observer at ground level. The horizontal leg is $80$ ft (distance to building base), the vertical leg is $h$ (building height), and the angle of elevation from the observer to the top is $42°$.

(b) The $42°$ angle is between the ground (adjacent = 80 ft) and the line of sight (hypotenuse). The building height is opposite:

$\tan 42° = \frac{h}{80}$

(c)

$h = 80 \tan 42° \approx 80(0.9004) \approx 72 \text{ feet}$

6. (a) The ships travel in perpendicular directions, forming a right triangle with legs $30$ and $40$ miles.

$d = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \text{ miles}$

The ships are 50 miles apart.

(b) At Ship B's position, the angle between the line to Ship A and the line back to port. Ship B is at the right-angle corner's opposite end: the angle at $B$ is between the east leg ($40$ mi, toward port) and the hypotenuse ($50$ mi, toward Ship A).

Using the sides at angle $B$: opposite to $B$'s angle is the north leg (Ship A's path) $= 30$ mi; adjacent (toward port from B) $= 40$ mi.

$\tan B = \frac{30}{40} = 0.75 \implies B = \tan^{-1}(0.75) \approx 37°$

The angle at Ship B is approximately $37°$.