Notes
Identities
- $\sin^2\theta + \cos^2\theta = 1$. Use it to find one function from the other, choosing the sign by quadrant.
- $\displaystyle \tan x = \frac{\sin x}{\cos x}$ and $1 + \tan^2 x = \sec^2 x$. To verify an identity, rewrite in sine and cosine and simplify one side. One counterexample shows an equation is not an identity.
Laws for any triangle
- Law of Sines: $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use for AAS, ASA, SSA. SSA can have 0, 1 or 2 triangles.
- Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$. Use for SAS and SSS.
Area
- Area $\displaystyle = \frac{1}{2}ab\sin C$ for two sides and the included angle.
- Heron's formula: $\displaystyle \sqrt{s(s - a)(s - b)(s - c)}$ with $s$ the semiperimeter.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle \cos^2\theta = 1 - \frac{25}{169} = \frac{144}{169}$, and cosine is negative in quadrant III.
$\displaystyle 1 + \tan^2 x = \sec^2 x = \frac{1}{\cos^2 x}$, so the product is 1.
$c^2 = 36 + 100 - 2(6)(10)\cos 120^\circ = 136 - 120(-0.5) = 196$.
$\displaystyle b = \frac{12\sin 65^\circ}{\sin 35^\circ} \approx \frac{12(0.9063)}{0.5736} \approx 19.0$.
$\displaystyle \frac{1}{2}(8)(9)\sin 30^\circ = 36(0.5) = 18$.
With SAS there is no known side-angle pair, so the Law of Sines cannot start. Use the Law of Cosines for the third side.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A triangular lot $ABC$ has $AB = 50$ m, $AC = 80$ m and angle $A = 60^\circ$.
(a) Find the length of side $BC$.
Show the answer and scoring guide
$BC^2 = 50^2 + 80^2 - 2(50)(80)\cos 60^\circ = 2500 + 6400 - 4000 = 4900$, so $BC = 70$ m.
Check off the points you earned:
(b) Find the area of the lot to the nearest square meter.
Show the answer and scoring guide
Area $\displaystyle = \frac{1}{2}(50)(80)\sin 60^\circ = 2000 \cdot \frac{\sqrt{3}}{2} \approx 1732$ square meters.
Check off the points you earned:
(c) Find angle $B$ to the nearest tenth of a degree.
Show the answer and scoring guide
$\displaystyle \cos B = \frac{AB^2 + BC^2 - AC^2}{2(AB)(BC)} = \frac{2500 + 4900 - 6400}{2(50)(70)} = \frac{1000}{7000} = \frac{1}{7}$, so $B \approx 81.8^\circ$.
Check off the points you earned: