Trigonometric Functions and the Unit Circle: math test solutions, Grade 12 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

a) \(210\cdot\dfrac{\pi}{180}=\dfrac{7\pi}{6}\) (1 pt)

b) \(\dfrac{11\cdot180}{4}=495^\circ\) (1 pt)

c) \(s=9\cdot\dfrac{2\pi}{3}=6\pi\approx18.85\text{ in}\) (1 pt)

d) \(A=\tfrac{1}{2}\cdot81\cdot\dfrac{2\pi}{3}=27\pi\approx84.82\text{ in}^{2}\) (1 pt)

a) Quadrant IV, reference \(\dfrac{\pi}{3}\): \(-\dfrac{\sqrt{3}}{2}\) (1 pt)

b) Quadrant II, reference \(\dfrac{\pi}{4}\): \(-\dfrac{\sqrt{2}}{2}\) (1 pt)

c) Quadrant III, reference \(\dfrac{\pi}{6}\), tangent positive: \(\dfrac{\sqrt{3}}{3}\) (1 pt)

d) \(\cos\dfrac{11\pi}{6}=\dfrac{\sqrt{3}}{2}\), so \(\sec=\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}\) (1 pt)

\(\sin^{2}\theta=1-\dfrac{64}{289}=\dfrac{225}{289}\); in quadrant III sine is negative, so \(\sin\theta=-\dfrac{15}{17}\) (1 pt).

\(\tan\theta=\dfrac{-15/17}{-8/17}=\dfrac{15}{8}\) (1 pt).

\(\csc\theta=-\dfrac{17}{15}\) (1 pt).

Sketch with the ladder as hypotenuse (1 pt).

Height \(=4.5\sin64^\circ\approx 4.04\text{ m}\) (1 pt).

Distance \(=4.5\cos64^\circ\approx 1.97\text{ m}\) (1 pt).

\(\cos\theta=-\dfrac{1}{2}\) (1 pt). The reference angle is \(\dfrac{\pi}{3}\) and cosine is negative in quadrants II and III.

\(\theta=\dfrac{2\pi}{3}\) (1 pt) or \(\theta=\dfrac{4\pi}{3}\) (1 pt).

\(\omega=15\cdot2\pi=30\pi\approx94.2\text{ rad/min}\) (1 pt).

\(v=120\cdot30\pi=3600\pi\approx11{,}309.7\text{ ft/min}\) (1 pt).

\(11{,}309.7\cdot\dfrac{60}{5280}\approx128.5\text{ mph}\) (1 pt).

Test yourself: quick challenge for Grade 12