Law of Cosines and Triangle Area

Develop included angles and side lengths, area from two sides and an angle.

Lesson 2 of 42 parts Practice Flashcards

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Included Angles and Side Lengths

Objective: Explain included angles and side lengths, apply the rule, and justify relevant restrictions.

For triangle sides a,b,c opposite A,B,C, the law of cosines is c²=a²+b²−2ab cos C. It applies to SAS data or to finding angles from SSS data. When C=90°, the cosine term vanishes and the formula becomes the Pythagorean theorem. The angle must be the included angle between the two sides in the cross term; length is the positive square root.

Formulas

$c²=a²+b²−2ab cos C$

Worked examples

Try itTwo sides are 21 and 28 with included angle 90°. Find the opposite side.Show answer

Step 1: Use d²=a²+b²−2ab cos C. Step 2: cos 90°=0, so d²=441+784=1225. Step 3: Take the positive square root: d=35.

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Included Angles and Side Lengths

Objective: Explain included angles and side lengths, apply the rule, and justify relevant restrictions.

Formulas

$c²=a²+b²−2ab cos C$

Worked examples

Try itTwo sides are 21 and 28 with included angle 90°. Find the opposite side.Show answer

Step 1: Use d²=a²+b²−2ab cos C. Step 2: cos 90°=0, so d²=441+784=1225. Step 3: Take the positive square root: d=35.

Area from Two Sides and an Angle

Objective: Explain area from two sides and an angle, apply the rule, and justify relevant restrictions.

Triangle area is (1/2)ab sin C when C is included between sides a and b. This follows by expressing a perpendicular height as b sin C and applying base times height divided by two. The formula works for both acute and obtuse included angles because sine is positive on (0,π). Area is reported in square units and differs from perimeter.

Formulas

$Area=(1/2)ab sin C$

Worked examples

Try itA triangle has sides 3 and 7 enclosing a 30° angle. Find its area.Show answer

Step 1: Use area=(1/2)ab sin C. Step 2: Substitute (1/2)(3)(7)(1/2). Step 3: The area is 21/4 square units.

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