Title: Example 1 Geometric Mean
Author: Tim Heppner
Last saved by: MHE
Company: MCGRAW-HILL

Geometry	Chapter 8

Lesson 8-7

Example 1  Two Sides and the Included Angle
Find a to the nearest tenth if c = 7, b = 4, and m∠A = 52.	
Use the Law of Cosines since the measures of two sides
and the included angle are known.

	a^2	= b^2 + c^2 – 2bc cos A	Law of Cosines
	a^2	= 4^2 + 7^2 - 2(4)(7) cos 52°	b = 4, c = 7, and m∠A = 52
	a^2	= 65 - 56 cos 52°	Simplify.
	a	= √(65 - 56 cos 52)	Take the square root of each side.
	a	≈ 5.5	Use a calculator.

Example 2  Three Sides
Find m∠S to the nearest degree.	

	s^2	= q^2 + r^2 - 2qr cos S	Law of Cosines
	15^2	= 21^2 + 17^2 - 2(21)(17) cos S	s = 15, q = 21, r = 17
	225	= 730 - 714 cos S	Simplify.
	-505	= -714 cos S	Subtract 730 from each side.
	-505/-714	= cos S	Divide each side by -714.
	S	= cos ^-1 	Solve for S.
	S	≈ 45.0	Use a calculator.

Example 3  Select a Strategy
Solve △UVW. Round angle measures to the 	
nearest degree and side measures to the nearest tenth.

We do not know whether △UVW a right triangle, so we must use the Law of Cosines or the Law of Sines. We know the measures of two sides and the included angle. This is SAS, so use the Law of Cosines. 

	u^2	= v^2 + w^2 – 2vw cos U	Law of Cosines
	u^2	= 13^2 + 12^2 - 2(13)(12) cos 41°	v = 13, w = 12, and m∠U = 41
	u	= 	Take the square root of each side.
	u	≈ 8.8	Use a calculator.

Next, we can find m∠V or m∠W. If we decide to find m∠V, we can use either the Law of Sines or the Law of Cosines to find this value. In this case, we will use the Law of Sines.

	(sin V)/v	= (sin U)/u	Law of Sines
	(sin V)/13	≈ (sin 41)/8.8	v = 12, u ≈ 8.8, and m∠U = 41
	8.8 sin V	≈ 13 sin 41°	Cross products
	sin V	≈ (13 sin 41°)/8.8	Divide each side by 8.8.
	V	≈ sin^–1 ﴾(13 sin 41°)/8.8﴿	Take the inverse sine of each side.
	V	≈ 76	Use a calculator.

Use the Angle Sum Theorem to find m∠W.

	m∠U + m∠V + m∠W	= 180	Angle Sum Theorem
	41 + 76 + m∠W	≈ 180	m∠U = 41 and m∠L ≈ 76
	m∠W	≈ 63	Subtract 117 from each side.

Therefore, u ≈ 14.2, m∠V ≈ 76, and m∠W ≈ 63.

Example 4  Use Law of Cosines to Solve Problems
REAL ESTATE  Mr. Sommers is buying some 	
property that is shaped like quadrilateral PQRS. 
Find the perimeter of the property.
Use the Pythagorean Theorem to find QS in △QRS.
	(QR)^2 + (RS)^2	= (QS)^2	Pythagorean Theorem	
	200^2 + 150^2	= (QS)^2	QR = 200, RS = 150
	62,500	= (QS)^2	Simplify.
	250	= QS	Take the square root of each side.

Next, use the Law of Cosines to find PQ in △PQS.
	(PQ)^2	= (PS)^2 + (QS)^2 - 2(PS)(QS) cos ∠PSQ	Law of Cosines
	(PQ)^2	= 225^2 + 250^2 - 2(225)(250) cos 45°	PS = 225, QS = 250, m∠PSQ = 45
	(PQ)^2	= 113,125 - 112,500 cos 45°	Simplify.
	PQ	= 	Take the square root of each side.
	PQ	≈ 183.2	Use a calculator.

The perimeter of the property is 150 + 200 + 183.2 + 225 or about 758.2 feet.