Law of Sines and Cosines Question 1

The figure below shows a map of Jennifer's neighborhood.

Which of the following expressions represents the straight-line distance, in miles, from Jennifer's home to school?

(Note: For a triangle with sides of length $a$ , $b$ , and $c$ that are opposite angles $\angle{A}$ , $\angle{B}$ , and $\angle{C}$ , respectively, $\displaystyle\frac{\sin{\angle{A}}}{a} = \frac{\sin{\angle{B}}}{b} = \frac{\sin{\angle{C}}}{c}$ and $c^2 = a^2+b^2-2ab\cos{\angle{C}}$ .)

\dfrac{4\sin{(130^\circ)}}{\sin{25^\circ}}

\dfrac{6\sin{(130^\circ)}}{\sin{25^\circ}}

\sqrt{4^2 + 6^2 - 2(4)(6)\cos{(25^\circ)}}

\sqrt{4^2 + 6^2 - 2(4)(6)\cos{(50^\circ)}}

\sqrt{4^2 + 6^2 - 2(4)(6)\cos{(130^\circ)}}

Approach

It is appropriate here to use law of cosines. We are trying to find the third side of the triangle give the other two sides. The angle opposite to this side is supplementary to $50^\circ$ .

a=4, b=6, \angle{C} = 130^\circ

c^2 = 4^2 + 6^2 - 2(4)(6)\cos{130^\circ}

c = \boxed{ \sqrt{4^2 + 6^2 - 2(4)(6)\cos{(130^\circ)}} }