Module 18Oblique Triangles

(Applications)

Florben G. Mendoza

FOUR CASES

CASE 1: One side and two angles are known (SAA or

ASA). Law of Sines

CASE 2: Two sides and the angle opposite one of them

are known (SSA). Law of Sines

CASE 3: Two sides and the included angle are known

(SAS). Law of Cosines

CASE 4: Three sides are known (SSS). Law of Cosines

Florben G. Mendoza

3

Law of Sines Law of Cosines

a

sin A=

b

sin B

c

sin B=

a2 = b2 + c2 – 2bc cos A

b2 = a2 + c2 – 2ac cos B

c2 = a2 + b2 – 2ab cos C

cos A = b2 + c2 - a2

2bc

cos B = a2 + c2 - b2

2ac

cos C = a2 + b2 - c2

2ab

Florben G. Mendoza

4

Practical applications of trigonometry often involve

determining distances that cannot be measured directly. In

many applications of trigonometry the essential problem is the

solution of triangles.

If enough sides and angles are known, the remaining sides and

angles as well as the area can be calculated, and the triangle is

then said to be solved. Problems

involving angles and distances in one plane are covered in this

lesson.

Florben G. Mendoza

5

Example 1: A navy aircraft is flying over a straight highway. When

the aircraft is in between the two cities that are 5 miles apart, he

determines that the angle of depression to two cities to be 32° and

48° respectively. Find the distance of the aircraft from the two

cities.

32° 48°

5 mi

32° 48°

A B

C

a = ?b = ? City BCity A

Florben G. Mendoza

6

32° 48°

5 mi

a = ?b = ?

A B

C

A = 32° B = 48° c = 5 mi

Given:

Find:a = ? b = ?

Step 1: ASA – Law of Sine

Step 2: C = 180° - (32° + 48°)

C = 180° - (A+B)

C = 100°

C = 180° - 80°

Step 3: c

sin C=

a

sin A

5

sin 100°=

a

sin 32°

a (sin 100°) = 5 (sin 32°)

sin 100° sin 100°

a = 2.69 mi

Florben G. Mendoza

7

32° 48°

5 mi

a = ?b = ?

A B

C

A = 32° B = 32° c = 5 mi

Given:

Find:a = ? b = ?

Step 4: c

sin C=

b

sin B

5

sin 100°=

b

sin 48°

b (sin 100°) = 5 (sin 48°)

sin 100° sin 100°b = 3.77 mi

Florben G. Mendoza

8

Example 2: Three circles of radii 100, 140, & 210 cm

respectively are tangent to each other externally. Find the

angles of the triangle formed by joining their centers.

A

C

B240

35031021

0 210

C

140

140

B

100

100

A ?

?

?

Florben G. Mendoza

9

A

C

B240

350310

Step 2: cos A = b2 + c2 - a2

2bc

cos A = (310)2 + (240)2 – (350)2

2(310)(240)

cos A = 31 200

148 800

cos A = 0.21

A = cos-1 0.21

A = 77.88°

Given:

a = 350

b = 310

c = 240

Find:

A = ?

B = ?

C = ?

Step 1: SSS – Law of Cosine

Florben G. Mendoza

10

Step 3: cos B = a2 + c2 - b2

2ac

cos B = (350)2 + (240)2 – (310)2

2(350)(240)

cos B = 161 000

217 000

cos B = 0.74

B = cos-1 0.74

B = 42.27°

A

C

B240

350310

Step 4:

C = 180° - (77.88° + 42.27°)

C = 180° - (A+B)

C = 59.85°

C = 180° - 120.15°

Florben G. Mendoza

11

Example 3: A pole casts a shadow of 15 meters long when the angle

of elevation of the sun is 61°. If the pole has leaned 15° from the

vertical directly towards the sun, find the length of the pole.

15 m

61°

15° ?

15 m

61° 105°

A

B

C

?

Florben G. Mendoza

12

105°61°

15 m

?

B

A C

Given:

A = 61°

C = 105°

b = 15 m

Find:

a = ?

Step 1: ASA – Law of Sine

Step 2: B = 180° - (61° + 105°)

B = 180° - (A + C)

B = 14°

B = 180° - 166°

Step 3: b

sin B=

a

sin A

15

sin 14°=

a

sin 61°

a (sin 14°) = 15 (sin 61°)

sin 14° sin 14°

a = 54.23 m

Florben G. Mendoza

13

Example 4: A tree on a hillside casts a shadow 215 ft down the

hill. If the angle of inclination of the hillside is 22° to the

horizontal and the angle of elevation of the sun is 52°, find the

height of the tree.

52 22 215 ft

22

30 38

22

68

112

38

215 ft

30

112

38

215 ft

A

C

B

?

Florben G. Mendoza

14

Given:

A = 30°

B = 112°

C = 38°

Find:

a = ?

A

B

C

30°

38°

112°

215 ft

?c

sin C=

a

sin A

215

sin 38°=

a

sin 30°

a (sin 38°) = 215 (sin 30°)

sin 38° sin 38°

a = 174.61 ft

c = 215 ft

Step 1:

Florben G. Mendoza

15

Example 5: A pilot sets out from an airport and heads in the

direction N 20° E, flying at 200 mi/h. After one hour, he makes a

course correction and heads in the direction N 50° E. Half an

hour after that, engine trouble forces him to make an emergency

landing. Find the distance between the airport & his final landing

point.

20°

50°

200

mi

100 mi

?

20°

40°

200

mi

100 mi

150°

?

A

B C

Florben G. Mendoza

16

200

mi

100 mi

150°

?

A

B C

Given:

B = 150°

a = 100 mi

c = 200 mi

Find:

b = ?

Step 2:

SAS – Law of CosineStep 1:

b2 = a2 + c2 – 2ac cos B

b2 = (100)2 + (200)2 – 2(100)(200) (cos 150°)

b2 = 50 000 – (- 34 641.02)

b2 = 84 641.02

b = 290.93

Florben G. Mendoza