Notes
Pythagorean Theorem
- $a^2 + b^2 = c^2$, with $c$ the hypotenuse. Common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
- Converse: with $c$ the longest side, $a^2 + b^2 = c^2$ is right, $>$ is acute, $<$ is obtuse.
Special right triangles
| Triangle | Side ratio |
|---|---|
| 45-45-90 | $\displaystyle 1 : 1 : \sqrt{2}$ |
| 30-60-90 | $\displaystyle 1 : \sqrt{3} : 2$ |
Similar right triangles
The altitude to the hypotenuse makes three similar triangles. Altitude $\displaystyle = \sqrt{xy}$ (the hypotenuse segments). Leg $\displaystyle = \sqrt{c \cdot x}$ (the adjacent segment).
Trigonometry
- $\displaystyle \sin A = \frac{\text{opp}}{\text{hyp}}$, $\displaystyle \cos A = \frac{\text{adj}}{\text{hyp}}$, $\displaystyle \tan A = \frac{\text{opp}}{\text{adj}}$.
- Use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ to find an angle.
- $\sin A = \cos(90^\circ - A)$.
- Angle of elevation looks up from horizontal; angle of depression looks down. They are equal for the same line of sight.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \sqrt{48^2 + 36^2} = \sqrt{2304 + 1296} = \sqrt{3600} = 60$.
The height splits it into two 30-60-90 triangles with short leg 5, so the height is $\displaystyle 5\sqrt{3}$.
The side adjacent to $\angle B$ is $BC = 24$, and the hypotenuse is 25.
$\displaystyle \tan^{-1}\left(\frac{3}{20}\right) \approx 8.5^\circ$. Using sine gives $8.6^\circ$, but 20 is the horizontal leg, not the hypotenuse.
The string is the hypotenuse: $150 \sin 55^\circ \approx 122.9$.
The hypotenuse is twice the short leg.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A drone hovers directly above a point on flat ground. An observer standing 200 meters from that point sees the drone at an angle of elevation of $32^\circ$. ($\tan 32^\circ \approx 0.6249$, $\cos 32^\circ \approx 0.8480$)
(a) Find the drone's height to the nearest tenth of a meter.
Show the answer and scoring guide
$\displaystyle \tan 32^\circ = \frac{h}{200}$, so $h = 200 \tan 32^\circ \approx 125.0$ m.
Check off the points you earned:
(b) Find the straight-line distance from the observer to the drone, to the nearest tenth.
Show the answer and scoring guide
$\displaystyle \cos 32^\circ = \frac{200}{d}$, so $\displaystyle d = \frac{200}{\cos 32^\circ} \approx 235.8$ m. (Check: $\displaystyle \sqrt{200^2 + 125^2} \approx 235.8$.)
Check off the points you earned:
(c) The drone rises straight up until it is 200 meters high. What is the new angle of elevation from the observer? Explain.
Show the answer and scoring guide
$\displaystyle \tan \theta = \frac{200}{200} = 1$, so $\theta = 45^\circ$. The legs are equal, so it is a 45-45-90 triangle.
Check off the points you earned: