oblique triangle
TRANSCRIPT
Oblique Triangle
An Oblique Triangle is a non-right triangle.
Oblique Triangle
β’ 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.
β’ 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 β Law of Sines
2. SSA β law of Sines ( ambiguous case)
3. SAS β 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
β’ 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
π
sin π΄=
π
sin π΅=
π
sin πΆ
Law of Sines
π
sin π΄=
π
sinπ΅π
sinπ΅=
π
sin πΆπ
sinπ΄=
π
sin πΆ
Law of Sines
1. Solve the triangle given.
Solution: A + B + C = 180
B = 180 β 51.2 β 48.6
B = 80.2o
From the law of sine23.5
sin 51.2=
π
sin 80.2
b=23.5(π ππ80.2)
π ππ51.2
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=
π
sin 48.6
c=23.5(π ππ48.6)
π ππ51.2
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=
π
sin 42
c=62.5(π ππ42)
π ππ112
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=
π
sin 26
b=62.5(π ππ26)
π ππ112
b= 29.5
Answers B=26o b= 45.1 c=29.5
If π πππ΅
π=
sin πΆ
π, then B =____
a) B = π ππβ1ππ πππΆ
πb) B = π ππβ1
ππ πππΆ
π
c)B = π ππβ1π
ππ ππ πΆ
Ans. a
If π
sin π΄=
π
sin π΅, then a =____
a) a =ππ πππ΅
sin π΄b) a =
π πππ΅
bsin π΄c) a =
ππ πππ΄
sin π΅
Ans. c
If π
sin π΄=
π
sin πΆ, then c =____
a)ππ πππ΄
sin πΆ= π b)
ππ πππΆ
sin π΄= π c)
π πππΆ
asin π΅= π
Ans. B
Oblique Triangle
Give the appropriate law.1. SAS β
Law of Cos2. SSS β
Law of cos
3. SSA βlaw of Sines ( ambiguous case)
4. ASA -Law of sin
5. SAA βlaw of sin
Law of Cosines
β’ For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 β 2bc cos A
b2 = a2 + c2 β 2ac cos B
c2 = a2 + b2 β 2ab cos C
Law of Cosines
β’ For any Triangle ABC with sides a,b, and c,
Use to solve the missing sides
a2 = b2 + c2 β 2bc cos A
b2 = a2 + c2 β 2ac cos B
c2 = a2 + b2 β 2ab cos C
Law of Cosines
β’ For any Triangle ABC with sides a,b, and c,
Use to solve the missing angles
cos A=b2 + c2 β a2
2ππ
cos B=a2 + c2 β b2
2ππ
cos C=a2 + b2 β c2
2ππ
Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
a2 = b2 + c2 β 2bc cos Aa2 = 12 + 32 β 2(1)(3) cos 80a2 = 1 + 9 β 6cos 80a2 = 10 β 1.04a2 = 8.96
a= 8.96a =2.99
a=2.99
Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
sin π΄
π=
sin π΅
πsin 80
2.99=
sin π΅
1(1 )sin 80
2.99= sinπ΅
0.3294 = sin Bπ ππβ1 0.3294 = π΅19.2o = B
a=2.99
19.2o
Law of Cosines
Solve the triangle with b=1, c=3, and A=80o.
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
β’ Solve the triangle with a = 5, b = 8, and c=9
a2 = b2 + c2 β 2bc cos A52 = 82 + 92 β 2(8)(9) cos A25=64+81-144cosA25=145-144cosA144cosA=145-25cosA=120/144cosA = 0.8333A= πππ β1 0.8333A = 33.6o
33.6o
Law of Cos
β’ Solve the triangle with a = 5, b = 8, and c=9
sin π΄
π=
sin π΅
πsin 33.6
5=
sin π΅
8(8 )sin 33.6
5= sinπ΅
0.8854 = sin Bπ ππβ1 0.8854 = π΅62.3o = B
33.6o62.3o
Law of Cos
β’ Solve the triangle with a = 5, b = 8, and c=9
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
β’ 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
β’ 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 β 32.5 β 54.98 = 92.52o
What is the length of side b?
b=3.08 C=79
β’ What is the size of Angle C?
C=40.51
β’ What is the size of Angle P?
The diagram shows part of a logo design.There is one known angle of 142Β°.Calculate the sizes of the other two angles.
β’ 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?
β’ What is the length of side c?
c2=5.32+3.62-2(5.3)(3.6)cos59c=4.63
β’ Find angle A
82=52+92-2(5)(9)cosAA=62.18
β’ Find angle B
β’ 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.