How to solve a triangle using the cosine and sine law 

Guest Jun 5, 2017

\(\underline{\text{Law of Sines}}\\ \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\\ \text{where the capital letters are the opposite}\\\text{angles of the corresponding sides.}\\ \underline{\text{Law of Cosines}}\\ a^2 + b^2 - 2ab\sin C = c^2\\ b^2 + c^2 - 2bc\sin A = a^2\\ a^2 + c^2 - 2ac\sin B = b^2\)

If AB = 16, BC = 8, AC = 12 and measure of angle A = 29 dg, solve the triangle.

First use law of sines to find the measure of angle C.

\(\dfrac{BC}{\sin A}=\dfrac{AB}{\sin C}\\ \dfrac{8}{\sin 29^{\circ}}=\dfrac{16}{\sin C}\\ \sin C = 2\sin29^{\circ}\\ C = \sin^{-1}(2\sin29^{\circ})\approx 75.8^{\circ}\)

Then use law of cosines to find the measure of angle B.(Actually I can use law of sines again but I just want to show you how to use law of cosines.)

\(BC^2+AB^2-2(BC)(AB)(\sin B)=AC^2\\ 8^2+16^2-2(8)(16)\sin B=12^2\\ 256\sin B = 176\\ \sin B = \dfrac{11}{16}\\ B = \sin^{-1}\left(\dfrac{11}{16}\right)\approx 43.4^{\circ}\)

MaxWong  Jun 6, 2017

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