H.Melikian/1200 7.1 The Sines Law Dr .Hayk Melikyan/ Departmen of Mathematics and CS/ [email protected] 1. Determining if the Law of Sines Can Be Used to Solve an Oblique Triangle 2. Using the Law of Sines to Solve the SAA Case or ASA Case 3. Using the Law of Sines to Solve the SSA Case 4. Using the Law of Sines to Solve Applied 5. Problems Involving Oblique Triangles
7.1 The Sines Law
Jan 17, 2016
Determining if the Law of Sines Can Be Used to Solve an Oblique Triangle Using the Law of Sines to Solve the SAA Case or ASA Case Using the Law of Sines to Solve the SSA Case Using the Law of Sines to Solve Applied Problems Involving Oblique Triangles. 7.1 The Sines Law. - PowerPoint PPT Presentation
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H.Melikian/1200 1
7.1 The Sines Law
Dr .Hayk Melikyan/ Departmen of Mathematics and CS/ [email protected]
1. Determining if the Law of Sines Can Be Used to Solve an Oblique Triangle
2. Using the Law of Sines to Solve the SAA Case or ASA Case
3. Using the Law of Sines to Solve the SSA Case
4. Using the Law of Sines to Solve Applied
5. Problems Involving Oblique Triangles
H.Melikian/1200 2
The Law of Sines
If A, B, and C are the measures of the angles of any triangle and if a, b, and c are the lengths of the sides opposite the corresponding angles, then
sin sin sin or
sin sin sin
a b c A B C
A B C a b c
H.Melikian/1200 3
The Law of Sines
The following information is needed to use the Law of Sines.
1. The measure of an angle must be known.
2. The length of the side opposite the known angle must be known.
3. At least one more side or one more angle must be known.
H.Melikian/1200 4
Determining of the Law of Sines Can Be Used to Solve an Oblique Triangle
Decide whether or not the Law of Sines can be used to solve each triangle. a. b. c.
A
B
C
a = 28
c = 54b = 40 50°
99°
A
BC
c = 14
a
b37°
47°
b = 15.4
B
A
Ca
c
No; We are not given the measure of any angle.
Yes; we are given an angle and the measure of its opposite side and an additional angle.
Yes; we are given an angle and the measure of its opposite side and an additional angle.
H.Melikian/1200 5
Solution We begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B.
A B C 180º The sum of the measurements of a triangle’s interior angles is 180º.
Add.83.5º B 180º
50º B 33.5º 180º
A = 50º and C = 33.5º.
B 96.5º Subtract 83.5º from both sides.
a
cA B
C
b 33.5º
50º
Text Example (ASA)
Solve triangle ABC if A 50º, C 33.5º, and b 76.
H.Melikian/1200 6
Solve triangle ABC if A 50º, C 33.5º, and b 76.
Solution Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b 76 and we found that B 96.5º. Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law of Sines to find a and c.
The solution is B 96.5º, a 59, and c 42.
Find a:
This is the known ratio.
a
cA B
C
b 33.5º
50º
Text Example cont.
asin A
bsin B
a
sin50 76
sin96.5
a 76sin 50
sin 96.5 59
csinC
bsin B
c
sin33.5 76
sin 96.5
c 76sin 33.5
sin 96.5 42
Find c:
H.Melikian/1200 7
Solving a SAA Triangle Using the Law of Sines
Solve the given oblique triangle. Round lengths to one decimal place.
50°
99°
A
BC
c = 14
a
b
Angles Sides
A = 99 a = ?
B = ?? b = ?
C = 50 c = 14
180
99 50 180
149 180
31
A B C
B
B
B
H.Melikian/1200 8
Solving a SAA Triangle Using the Law of SinesSolve the given oblique triangle. Round lengths to one decimal place. Find a: Find b:
50°
99°
A
BC
c = 14
a
b
Angles Sides
A = 99 a = ?
B = 31 b = ?
C = 50 c = 14
sin sin14
sin99 sin5014sin99
sin5018.1
a c
A Ca
a
a
sin sin14
sin31 sin5014sin31
sin509.4
b c
B Cb
b
b
H.Melikian/1200 9
Solving a ASA Triangle Using the Law of Sines
Solve oblique triangle ABC if B = 42, A = 57, and c = 18.6 cm.
Angles Sides
A = 57 a = ?
B = 42 b = ?
C = c = 18.6
sin sin18.6
sin57 sin8118.6sin57
sin8115.8
a c
A Ca
a
a
42°
57°
b
A
Ca
c = 18.6
B
180 57 42 81C
81
sin sin18.6
sin 42 sin8118.6sin 42
sin8112.6
b c
B Cb
a
a
H.Melikian/1200 10
Two Triangles
No Triangle
b a
A
h = b sin A
a is less than h and not long
enough to form a triangle.
b
A
a h = b sin A
a = h and is just the right
length to form a right triangle.
a ab
A
h = b sin A
a is greater than h and a is less than b. Two distinct triangles are formed.
b a
A
a is greater than h and a is greater
than b. One triangle is formed.
The Ambiguous Case (SSA)
One Right Triangle
One Triangle
H.Melikian/1200 11
H.Melikian/1200 12
Solving a SSA Triangle Using the Law of Sines ( Quiz No)
Solve the triangle; if possible.
Angles Sides
A = ? a = 20
B = ? b =
C = 50 c = 10
sin sin
sin sin50
20 1020sin50
sin10
sin 1.5
A C
a cA
A
A
Since there is no angle A for which sin A > 1, there can be no triangle with the given measurements.
H.Melikian/1200 13
Solving a SSA Triangle Using the Law of Sines (One Triangle)
Solve the triangle; if possible.
Angles Sides
A = 40 a = 30
B = ? b = 20
C = ? c = ?
sin sin
sin 40 sin
30 2020sin 40
sin30
sin 0.4285
A B
a bB
A
B
Two possible angles:
25.4 or 180 25.4 154.6
154.6 is too large when combined with the given angle.
H.Melikian/1200 14
Solving a SSA Triangle Using the Law of Sines (One Triangle)--cont
Solve the triangle; if possible.
Angles Sides
A = 40 a = 30
B = 25.4 b = 20
C = 114.6 c = ?
180 40 25.4 114.6C
sin sin
sin 40 sin114.6
3030sin114.6
sin 4042.3
A C
a c
c
c
c
H.Melikian/1200 15
Solving a SSA Triangle Using the Law of Sines (Two Triangles)Solve the triangle; if possible.
Angles Sides
A = 35 a = 60
B = ? b = 80
C = ? c = ?
sin sin
sin35 sin
60 8080sin35
sin60
sin 0.7648
A B
a bB
B
B
Two possible angles:
49.9 or 180 49.9 130.1
There are two possible triangles.
H.Melikian/1200 16
Solving a SSA Triangle Using the Law of Sines (Two Triangles)
Solve the triangle; if possible.
Angles Sides Angles Sides
A = 35 a = 60 A = 35 a = 60
B = 49.9
b = 80 B = 130.1
b = 80
C = c = C = c =
sin sin
sin35 sin95.1
6060sin95.1
sin35104.2
A C
a c
c
c
c
180 35 49.9 95.1
180 35 130.1 14.9
C
or
C
95.1 14.9104.2
H.Melikian/1200 17
Solving a SSA Triangle Using the Law of Sines (Two Triangles)
Solve the triangle; if possible.
Angles Sides Angles Sides
A = 35 a = 60 A = 35 a = 60
B = 49.9
b = 80 B = 130.1
b = 80
C = c = C = c =
sin sin
sin35 sin14.9
6060sin14.9
sin3526.9
A C
a c
c
c
c
95.1 14.9104.2 26.9
H.Melikian/1200 18
Solving an Applied Problem
Martin wants to measure the distance across a river. He has made a sketch. Find the distance across the river, a.
126°
23°
348 feet
a
180 23 126 31C
31
sin sin348
sin 23 sin31348sin 23
sin31264 ft
a b
A Ba
a
a
H.Melikian/1200 19
Determining the Distance a Ship is from Port
A ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E.
Draw a diagram.Find angle A.
74 53 21A
H.Melikian/1200 20
Determining the Distance a Ship is from Port-contA ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E.
sin sin
sin sin 21
63 6963sin 21
sin69
sin 0.3272
C A
c aC
A
A
Angles Sides
A = 21 a = 69
B = b =
C = c = 63
Two possible angles:19.1 or 180 19.1 160.9 160.9 will not work
H.Melikian/1200 21
Determining the Distance a Ship is from Port-contA ship set sail from port at a bearing of N 53E and sailed 63 km to point B. The ship then turned and sailed an additional 69 km to point C. Determine the distance from port to point C if the ship’s final bearing is N 74E.
sin sin69
sin139.9 sin 2169sin139.9
sin 21124.0
b a
B Ab
b
b
Angles Sides
A = 21 a = 69
B = 139.9
b =
C = 19.1 c = 63
The ship is 124 miles from port to point C.
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