Notes
Laws for any triangle
- Law of Sines: $\displaystyle \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$. Use for AAS, ASA, and SSA.
- Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$. Use for SAS and SSS.
- Ambiguous case (SSA): with $A$ acute, let $h = b\sin A$. Then $a < h$ gives no triangle, $a = h$ one, $h < a < b$ two, and $a \ge b$ one. With $A$ obtuse there is one triangle only if $a > b$.
Area
- $\displaystyle \text{Area} = \frac{1}{2}ab\sin C$, or Heron's formula $\displaystyle \sqrt{s(s-a)(s-b)(s-c)}$ with $\displaystyle s = \frac{a+b+c}{2}$.
Vectors
- $\vec{v} = \langle x, y \rangle$ has magnitude $\displaystyle \sqrt{x^2 + y^2}$ and components $\langle |\vec{v}|\cos\theta, |\vec{v}|\sin\theta \rangle$.
- Add and scale component by component. The dot product $ac + bd$ is 0 for perpendicular vectors.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$\displaystyle b = \frac{10\sin 30^\circ}{\sin 45^\circ} = \frac{5}{\sqrt{2}/2} = 5\sqrt{2}$.
When $A$ is obtuse, side $a$ must be longer than $b$. Here $5 < 8$, so no triangle exists.
$c^2 = 25 + 36 - 2(5)(6)\cos 120^\circ = 61 + 30 = 91$.
$\displaystyle \frac{1}{2}(8)(9)\sin 150^\circ = 36 \cdot \frac{1}{2} = 18$.
$8\cos 120^\circ = -4$ and $\displaystyle 8\sin 120^\circ = 4\sqrt{3}$.
$1 \cdot 3 + 2 \cdot (-4) = 3 - 8 = -5$. The dot product is a number, not a vector.
Free response
Write your answer first, then open the scoring guide.
Free-response question
In triangle $ABC$, side $a = 7$, side $b = 8$, and the angle between them is $C = 60^\circ$.
(a) Find side $c$ using the Law of Cosines.
Show the answer and scoring guide
$c^2 = 7^2 + 8^2 - 2(7)(8)\cos 60^\circ = 49 + 64 - 56 = 57$, so $\displaystyle c = \sqrt{57} \approx 7.55$.
Check off the points you earned:
(b) Find the exact area of the triangle.
Show the answer and scoring guide
Area $\displaystyle = \frac{1}{2}(7)(8)\sin 60^\circ = 28 \cdot \frac{\sqrt{3}}{2} = 14\sqrt{3} \approx 24.2$.
Check off the points you earned:
(c) Find angle $B$ (opposite side $b = 8$, the longest side) to the nearest tenth of a degree.
Show the answer and scoring guide
$\displaystyle \cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{49 + 57 - 64}{2 \cdot 7 \cdot \sqrt{57}} = \frac{42}{14\sqrt{57}} = \frac{3}{\sqrt{57}} \approx 0.397$, so $B \approx 66.6^\circ$.
Check off the points you earned: