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Math I

Laws of Sines and Cosines and Area: Practice and Methods

Knowing some of the sides and angles of a triangle determines the rest. Choose the law that fits the information you are given.

Basic, Standard: Grade 10 · Term 3 / Advanced: Grade 12 · Exam prep

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The laws of sines and cosines

Law of sines (RR is the circumradius)
asin⁡A=bsin⁡B=csin⁡C=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R
Law of cosines
a2=b2+c2−2bccos⁡A,cos⁡A=b2+c2−a22bca^2 = b^2 + c^2 - 2bc\cos A,\qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}

Use the law of sines for "one side and two angles" or circumradius problems, and the law of cosines for "two sides and the included angle" or "three sides". It also follows that sin⁡A:sin⁡B:sin⁡C=a:b:c\sin A : \sin B : \sin C = a : b : c.

Area and the incircle

Area
S=12bcsin⁡A,S=12r(a+b+c)(r is the inradius)S = \frac{1}{2}bc\sin A,\qquad S = \frac{1}{2}r(a + b + c)\quad (r \text{ is the inradius})

Given three sides, find cos⁡A\cos A by the law of cosines, then sin⁡A\sin A from sin⁡2A+cos⁡2A=1\sin^2 A + \cos^2 A = 1, then the area. Example: for a=7a = 7, b=5b = 5, c=3c = 3, cos⁡A=−12\cos A = -\frac{1}{2}, so A=120∘A = 120^\circ and S=1534S = \frac{15\sqrt{3}}{4}.

Two solutions, cyclic quadrilaterals, surveying

Given aa, bb and AA, the law of cosines becomes a quadratic equation in cc, which may have two positive solutions; every positive solution is an answer.

Cyclic quadrilaterals

Opposite angles add to 180∘180^\circ, so cos⁡D=−cos⁡B\cos D = -\cos B and sin⁡D=sin⁡B\sin D = \sin B. Express diagonal AC in two ways using the law of cosines in △ABC\triangle ABC and △ACD\triangle ACD to find cos⁡B\cos B.

In surveying problems, first use the law of sines in a triangle on the ground to find a distance, then use tan⁡\tan in the vertical right triangle to find the height.