Solving oblique (non-right) triangles

• Suppose the triangle to be solved does not contain a right angle. If we are given three parts of the triangle, not all angles, then we can solve for the remaining parts.

• Usually we label the angles of the triangle as A, B, C and the sides opposite these angles as a, b, and c, respectively.

• The Law of Sines given by

can be used to solve the following two cases.

1. Two angles and any side (AAS and ASA)

2. Two sides and an angle opposite one of them (SSA)

b a

cA

C

B

or C sin

c

B sin

b

A sin

a

c

Csin

b

Bsin

a

Asin

An ASA example using the Law of Sines

• Two fire stations are located 56.7 miles apart, at points A and B. There is a forest fire at point C. If CAB = 54° and CBA = 58°, which fire station is closer? How much closer?

• The fire station at point B is closer to the fire by 51.861 – 49.474 = 2.387 miles.

54°

68°

58°

C

A B

b

c = 56.7 miles

a

miles. 49.474 68sin

54sin 56.7 a so ,

54sin

a

68sin

7.56

miles. 51.861 68sin

58sin 56.7 b so ,

58sin

b

68sin

7.56

The ambiguous case for solving a triangle (SSA)• Suppose we are given two sides of a triangle and the angle

opposite one of them, and we are asked to find the remaining parts of the triangle. Depending on the data given, there may be: two possible triangles, one possible triangle, or no possible triangle.

• Example. CAB = 45°, b = 1, and various values of a:

No triangle exists.

b=145°

a=2/2

b=1 a=3/10

One triangle exists.A

b=145°

a=8/10

A

Two triangles exist.

b=145°

A

a=11/10

45°A

C

C C

B B

C

The area of an oblique triangle

• The area of a triangle is given by the formula

• For example, in the triangle below,

• By choosing other sides of the triangle as the base, we obtain: The area of any triangle is one-half the product of the lengths of two sides times the sine of the included angle. That is,

b a

cA

C

B

ht).base)(heig( Area 21

A.sin bcA)sin c)(b(ht)base)(heig( Area 21

21

21

B.sin acCsin abAsin bc Area 21

21

21

Solving oblique triangles, continued

b a

cA

C

B

• Two cases remain for solving an oblique triangle.

3. Three sides (SSS)

4. Two sides and their included angle (SAS)

The Law of Cosines (given below) is used for these cases.

A cos2bccba 222

B cos2accab 222

C cos2abbac 222

2bc

acbA cos

222

Alternative FormStandard Form

2ac

bcaB cos

222

2ab

cbaC cos

222

An SAS example using the Law of Cosines

• A person leaves her home and walks 5 miles due east and then 3 miles northeast. How far away from home (as the crow flies) is she? As seen from her home, what is the angle between the easterly direction and the direction to her destination?

• By applying the Law of Cosines, we can solve for x:

• A application of the alternative form gives:

Home

Destination

x

5

345°135°θ

N

. 55.2 )2/2(30 34 135 cos352 3 5 x 222 miles. 7.43 55.2 xSo

so 0.9583,7.43)))(5((2/)37.43 (5 θ cos 222 . 16.6 583)arccos(0.9 θ

An SAS example using the Law of Cosines, continued

Instead of applying the alternative form of the Law of Cosines to find θ, we could have used the Law of

Sines, which is an easier computation.

Home

Destination

7.43

5

345°135°θ

N

so ,2855.043.7

135sin3θsin

43.7

135sin

3

θsin

before. as ,6.162855.0arcsinθ

An SSS example using the Law of Cosines• Find the angles of the triangle.

• Find the angle opposite the longest side first since it will be the largest angle. This will reduce uncertainty.

• Since cos B is negative, B is the triangle's only obtuse angle. B = arccos(–0.45089) = 116.80º. Use the Law of Sines to determine A.

• C =

8 14

19

B

AC

.45089.014)))(8((2/)1914 (8 B cos 222

08.22)37583.0arcsin(A

37583.019

116.80sin 8 A sin

.12.4180.11608.22180

Heron's area formula

• The Law of Cosines can be used to establish another formula for the area of a triangle.

• Heron's Area Formula Given any triangle with sides of lengths a, b, and c, the area of the triangle is

• Example. Using Heron's formula, find the area of an equilateral triangle with sides of length 1.

In this example, s = 3/2 and s – a = s – b = s – c = 1/2.

.2c)b(a s where

c)b)(sa)(ss(sArea

43

21

21

21

23 ))()((Area

Solve the triangle--An SSA example.

• Given the triangle below (not to scale).

• It follows that sin = ______ , = ______, and φ= ______. The area of the given triangle is ______ cm2. The length x = _____ cm.

110

10cm3cm

x

φ

How should we solve?

• Determine the method needed to solve these triangles. The triangles are not to scale.

12

18

16

(a) (b)

55º

35º

(c)

115º

14.5

9

18

y z

A B

C

θ

z

θ

φ

(d)

20ºθ

φ

10

8z