Sign in

Precalculus

Unit 6

Applications of Trigonometry

6 topics · 24 practice questions · Free responseNot yet reviewed

Notes

Solving triangles

  • Law of Sines: $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. Use it with AAS, ASA and SSA. SSA can be ambiguous (0, 1 or 2 triangles).
  • Law of Cosines: $c^2 = a^2 + b^2 - 2ab\cos C$. Use it with SAS and SSS.
  • Area: $\displaystyle \frac{1}{2}ab\sin C$, or Heron's formula $\displaystyle \sqrt{s(s-a)(s-b)(s-c)}$.

Vectors

  • $\displaystyle |\langle a, b \rangle| = \sqrt{a^2 + b^2}$. A vector of magnitude $r$ at angle $\theta$ is $\langle r\cos\theta, r\sin\theta \rangle$.
  • Dot product: $\langle a, b \rangle \cdot \langle c, d \rangle = ac + bd$. It is 0 for perpendicular vectors, and $\displaystyle \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}$.

Topics in this unit

Each topic has its own short notes and practice.

Practice

In triangle $ABC$, $a = 5$, $b = 5$ and $C = 120^\circ$. What is $c$?
A triangle has $A = 60^\circ$, $B = 45^\circ$ and $a = 6$. What is $b$?
A triangle has sides 13, 14 and 15. Heron's formula gives what area?
What is the magnitude of $\langle 8, -6 \rangle$?
What is $\langle 3, -2 \rangle \cdot \langle 4, 5 \rangle$?
A triangle has two sides of 4 and 6 with an included angle of $150^\circ$. What is its area?

Free response

Write your answer first, then open the scoring guide.

Free-response question

Triangle $ABC$ has sides $a = 7$, $b = 8$ and $c = 9$.

(a) Find $\cos C$ (the angle opposite side $c$).

Show the answer and scoring guide

$\displaystyle \cos C = \frac{49 + 64 - 81}{2 \cdot 7 \cdot 8} = \frac{32}{112} = \frac{2}{7}$.

Check off the points you earned:

(b) Find the area using Heron's formula.

Show the answer and scoring guide

$s = 12$, so area $\displaystyle = \sqrt{12 \cdot 5 \cdot 4 \cdot 3} = \sqrt{720} = 12\sqrt{5}$.

Check off the points you earned:

(c) Confirm the area using $\frac{1}{2}ab\sin C$.

Show the answer and scoring guide

$\displaystyle \sin C = \sqrt{1 - \frac{4}{49}} = \frac{\sqrt{45}}{7} = \frac{3\sqrt{5}}{7}$. Area $\displaystyle = \frac{1}{2}(7)(8)\cdot\frac{3\sqrt{5}}{7} = 12\sqrt{5}$.

Check off the points you earned: