Oblique Triangle

Oblique Triangle

An Oblique Triangle is a non-right triangle.

Oblique Triangle

• There are several laws that can be use to solve oblique triangle. These are the law of sines, law of cosines and law of tangents.

• As in solving right triangles, you should know three parts of an oblique triangle to find the other three missing parts.

Oblique Triangle

Four Cases

1. ASA or SAA – Law of Sines

2. SSA – law of Sines ( ambiguous case)

3. SAS – Law of cosines

4. SSS- Law of Cosines

Oblique Triangle

1. Given: A, b, C Law of Sin

2. c, B, C Law of Sin

3. c, a, C Law of Sin

4. b,A,c Law of Cos

5. c, B, a Law of cos

6. a, b, c Law of cos

Oblique Triangle

Law of cos

Law of sin

Law of cos

Law of Sines

• If A, B, and C, are the angles of any triangle, and a,b, and c, are respectively, the measures of the sides opposite these angles, then

𝑎

sin 𝐴=

𝑏

sin 𝐵=

𝑐

sin 𝐶

Law of Sines

𝑎

sin 𝐴=

𝑏

sin𝐵𝑏

sin𝐵=

𝑐

sin 𝐶𝑎

sin𝐴=

𝑐

sin 𝐶

Law of Sines

1. Solve the triangle given.

Solution: A + B + C = 180

B = 180 – 51.2 – 48.6

B = 80.2o

From the law of sine23.5

sin 51.2=

𝑏

sin 80.2

b=23.5(𝑠𝑖𝑛80.2)

𝑠𝑖𝑛51.2

b= 29.7

80.2o

29.7

Law of Sines

1. Solve the triangle given.

B= 80.2

b = 29.7

From the law of sine23.5

sin 51.2=

𝑐

sin 48.6

c=23.5(𝑠𝑖𝑛48.6)

𝑠𝑖𝑛51.2

c= 22.6

22.6

80.2o

29.7

Solve the triangle ABC, given a=62.5, A=112o, and C=42

B=180-112-42

B=2662.5

sin 112=

𝑐

sin 42

c=62.5(𝑠𝑖𝑛42)

𝑠𝑖𝑛112

c= 45.1

Solve the triangle ABC, given a=62.5, A=112o, and C=42

B=180-112-42

B=2662.5

sin 112=

𝑏

sin 26

b=62.5(𝑠𝑖𝑛26)

𝑠𝑖𝑛112

b= 29.5

Answers B=26o b= 45.1 c=29.5

If 𝑠𝑖𝑛𝐵

𝑏=

sin 𝐶

𝑐, then B =____

a) B = 𝑠𝑖𝑛−1𝑏𝑠𝑖𝑛𝐶

𝑐b) B = 𝑠𝑖𝑛−1

𝑐𝑠𝑖𝑛𝐶

𝑏

c)B = 𝑠𝑖𝑛−1𝑏

𝑐𝑠𝑖𝑛 𝐶

Ans. a

If 𝑎

sin 𝐴=

𝑏

sin 𝐵, then a =____

a) a =𝑏𝑠𝑖𝑛𝐵

sin 𝐴b) a =

𝑠𝑖𝑛𝐵

bsin 𝐴c) a =

𝑏𝑠𝑖𝑛𝐴

sin 𝐵

Ans. c

If 𝑎

sin 𝐴=

𝑐

sin 𝐶, then c =____

a)𝑎𝑠𝑖𝑛𝐴

sin 𝐶= 𝑐 b)

𝑎𝑠𝑖𝑛𝐶

sin 𝐴= 𝑐 c)

𝑠𝑖𝑛𝐶

asin 𝐵= 𝑐

Ans. B

Oblique Triangle

Give the appropriate law.1. SAS –

Law of Cos2. SSS –

Law of cos

3. SSA –law of Sines ( ambiguous case)

4. ASA -Law of sin

5. SAA –law of sin

Law of Cosines

• For any Triangle ABC with sides a,b, and c,

Use to solve the missing sides

a2 = b2 + c2 – 2bc cos A

b2 = a2 + c2 – 2ac cos B

c2 = a2 + b2 – 2ab cos C

Law of Cosines


Use to solve the missing sides

a2 = b2 + c2 – 2bc cos A

b2 = a2 + c2 – 2ac cos B

c2 = a2 + b2 – 2ab cos C

Law of Cosines


Use to solve the missing angles

cos A=b2 + c2 − a2

2𝑏𝑐

cos B=a2 + c2 − b2

2𝑎𝑐

cos C=a2 + b2 − c2

2𝑎𝑏

Law of Cosines

Solve the triangle with b=1, c=3, and A=80o.

a2 = b2 + c2 – 2bc cos Aa2 = 12 + 32 – 2(1)(3) cos 80a2 = 1 + 9 – 6cos 80a2 = 10 – 1.04a2 = 8.96

a= 8.96a =2.99

a=2.99

Law of Cosines


sin 𝐴

𝑎=

sin 𝐵

𝑏sin 80

2.99=

sin 𝐵

1(1 )sin 80

2.99= sin𝐵

0.3294 = sin B𝑠𝑖𝑛−1 0.3294 = 𝐵19.2o = B

a=2.99

19.2o

Law of Cosines


Since A+B+C = 180o

Then 80o +19.2o +C=180o

99.2o +C =180o

C =180o -99.2o

C = 80.8o

a=2.99

19.2o

80.8o

Law of

• Solve the triangle with a = 5, b = 8, and c=9

a2 = b2 + c2 – 2bc cos A52 = 82 + 92 – 2(8)(9) cos A25=64+81-144cosA25=145-144cosA144cosA=145-25cosA=120/144cosA = 0.8333A= 𝑐𝑜𝑠−1 0.8333A = 33.6o

33.6o

Law of Cos


sin 𝐴

𝑎=

sin 𝐵

𝑏sin 33.6

5=

sin 𝐵

8(8 )sin 33.6

5= sin𝐵

0.8854 = sin B𝑠𝑖𝑛−1 0.8854 = 𝐵62.3o = B

33.6o62.3o

Law of Cos


Since A+B+C = 180o

33.6o +62.3o +C=180o

95.9o +C =180o

C =180o -95.9o

C = 84.1o33.6o62.3o

84.1o

• A Triangular lot has dimensions 20.6m, 31.4m, and 38.3m. Find the angles at the corners of the property.

20.62=31.42+38.32 -2(31.4)(38.3)cosA

424.36=2452.85-2405.24cosA

cosA=2028.49/2405.24

cosA=0.8434

A= 32.5o

• A Triangular lot has dimensions 20.6m, 31.4m, and 38.3m. Find the angles at the corners of the property.

sin 32.5/20.6 = sin B/31.4

0.8190 =sin B

B = 54. 98o or 55

C =180 – 32.5 – 54.98 = 92.52o

What is the length of side b?

b=3.08 C=79

• What is the size of Angle C?

C=40.51

• What is the size of Angle P?

The diagram shows part of a logo design.There is one known angle of 142°.Calculate the sizes of the other two angles.

• Mrs Jones goes on a round trip from Town A to Town B to Town C and back to Town A, as shown in the following diagram. All roads are straight. To the nearest mile. How long is the round trip?

• What is the length of side c?

c2=5.32+3.62-2(5.3)(3.6)cos59c=4.63

• Find angle A

82=52+92-2(5)(9)cosAA=62.18

• Find angle B

• Ayton is 25 miles due north of Beeton. Ceeton lies to the east side the road joining Ayton to Beeton, and is 47 miles from Ayton and 63 miles from Beeton. (All roads are straight.)Calculate the three-figure bearing of Ceeton from Ayton.Note A three-figure bearing is always measured in a clockwise sense from the direction North.

Education

Oblique Triangle