Notes
Ratios and special triangles
- $\displaystyle \sin = \frac{\text{opp}}{\text{hyp}}$, $\displaystyle \cos = \frac{\text{adj}}{\text{hyp}}$, $\displaystyle \tan = \frac{\text{opp}}{\text{adj}}$. Their reciprocals are $\csc$, $\sec$ and $\cot$.
- 45-45-90: sides $\displaystyle x, x, x\sqrt{2}$. 30-60-90: sides $\displaystyle x, x\sqrt{3}, 2x$. Use them for exact values.
- Cofunctions: $\sin\theta = \cos(90^\circ - \theta)$.
Solving right triangles
- To find a side, choose the ratio that links the known angle and sides. To find an angle, use $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$.
- The acute angles add to $90^\circ$. Use the Pythagorean theorem when two sides are known.
Applications
- Angle of elevation is measured up from the horizontal; angle of depression is measured down. They are equal for the two ends of one line of sight.
- Draw and label a right triangle before calculating.
Topics in this unit
Each topic has its own short notes and practice.
Practice
Tangent is opposite over adjacent.
In a 30-60-90 triangle, $\displaystyle \tan 60^\circ = \frac{x\sqrt{3}}{x}$.
Hypotenuse $\displaystyle = 9\sqrt{2} \cdot \sqrt{2} = 18$.
The rise is opposite the angle: $12\sin 15^\circ \approx 3.1$.
$100\sin 55^\circ \approx 81.9$.
The opposite side is 3 (3-4-5 triangle), so $\displaystyle \sin\theta = \frac{3}{5}$ and $\displaystyle \csc\theta = \frac{5}{3}$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A surveyor stands 120 m from the base of a building. Her eye is 1.6 m above the ground, and the angle of elevation to the top of the building is $32^\circ$.
(a) How far is the top of the building above her eye level? Round to the nearest tenth of a meter.
Show the answer and scoring guide
$\displaystyle \tan 32^\circ = \frac{h}{120}$, so $h = 120\tan 32^\circ \approx 75.0$ m.
Check off the points you earned:
(b) What is the height of the building?
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Add her eye height: $75.0 + 1.6 = 76.6$ m.
Check off the points you earned:
(c) She moves to a point 90 m from the building (same eye height). Find the new angle of elevation to the nearest tenth of a degree.
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$\displaystyle \tan\theta = \frac{75.0}{90} \approx 0.833$, so $\theta = \tan^{-1}(0.833) \approx 39.8^\circ$.
Check off the points you earned: