Week 17 – The Sine Law and Cosine Law

Near the end of this Semester, we learned to solve triangles that don’t involve right angles. At first to me, it was really frustrating that I can’t use the Pythagoras Theorem since there were no right angles, but after learning the 2 trigonometry laws mentioned on the title, I felt like that helped me a lot.

For the Sine Law, \frac{Sin A}{a}=\frac{Sin B}{b}=\frac{Sin C}{c} or \frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}, where the capital letters are angles and the small letters represent the opposite side of the coressponding angle.

For the Cosine Law, a^2=b^2+c^2-2(b)(c)(Cos A), P.S., all variables and unknowns can be switched around, such as like this: Cos A=\frac{b^2+c^2-a^2}{2(b)(c)}. All letter rules mentioned in the Sine Law also apply to this.

Here is a question that uses both Laws: In \Delta RST, ∠R=64 °, RS=8.3m, and RT=9.0m. Solve the triangle.

Form the triangle I’ve drawn, I instantly figured out that I’d have to use Cosine law to figure out ST, so ST^2=9.0^2+8.3^2-2(9.0)(8.3)(Cos 64), ST \doteq 9.2 M.

Using the answer, I used the Sine Law to find both ∠S and ∠T:

\frac{Sin 64}{9.2}=\frac{Sin S}{9.0}, S \doteq 62 °

\frac{Sin 64}{9.2}=\frac{Sin T}{8.3}, T \doteq 54 °

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