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Law of Sines
• Given the triangle below …
… the law of sines is given by …
sin sin sin
a b c
Law of Sines• Note that in each ratio, the sine of the angle is
written over the length of the side opposite that angle.
sin sin sin
a b c
• Note also that the triangle is not a right triangle, so the pythagorean theorem cannot be used.
Example 1:Solve the triangle with the given measures:
30 , 110 , 10 b cm
180
• Since the three angles of a triangle add up to 180 degrees …
30 110 180
40
30 , 110 , 10 b cm
sin sin
a b
• Now use the law of sines. Since side b is given, one ratio will include side b and angle β (the angle opposite side b).
• The other ratio is our choice since we know the value of both angles.
sin 30 sin110
10
a
sin 30 sin110
10
a
sin110 10sin 30a
10sin 30
sin110a
5.3a cm
• Note: make sure the calculator is set to degree mode.
30 , 110 , 10 b cm
sin sin
c b
• Now find the last side c. Use side b for the other ratio since it is given. Using the rounded value of a would lead to further rounding error.
sin 40 sin110
10
c
5.3a40
• All missing measures have been found and the triangle is solved.
sin 40 sin110
10
c
sin110 10sin 40c
10sin 40
sin110c
6.8c cm
Example 2:Solve the triangle with the given measures:
52 , 6 , 5 a in b in
• Since side b and angle β are both given, use them for one ratio.
• Since side a is given, use it for the other ratio.
sin sin
a b
52 , 6 , 5 a in b in
sin sin 52
6 5
5sin 6sin 52
6sin52sin
5
1 6sin52sin5
71
6sin52sin
5
• Note: make sure the calculator is set to degree mode.
• We know two angles now, so find the third angle.
52 , 6 , 5 a in b in
71
180
71 52 180
57
• Find the remaining side.
52 , 6 , 5 a in b in
71 57
sin sin
c b
sin 57 sin 52
5
c
sin 57 sin 52
5
c
sin 52 5sin 57 c
5sin57
sin52c
5.3c in
• All missing measures have been found and the triangle is solved.
• Before we move on, consider one of the calculations that was made in this problem.
1 6sin52sin5
71
• The calculator gave us a value of …
1sin 0.9456
• Consider a unit circle with the given information.
• There is another possible value for the angle. Using a reference angle of 71 degrees …
… we find another angle that will also solve the equation.
180 71 109
• Note that 109 degrees is possible for an angle in a triangle.
• Recall the sides and angles that have been found up to this point
52 , 6 , 5 a in b in• Given:
• Determined:
71
57
5.3c in
• Now consider a second possible triangle.
• Use the second value found for α and re-solve the triangle.
52 , 6 , 5 a in b in
109
180
109 52 180
19
52 , 6 , 5 a in b in
109 19
sin sin
c b
sin19 sin 52
5
c
sin 52 5sin19 c
sin 52 5sin19 c
5sin19
sin52c
2.1c in
• The given information led to two different triangles.
71
57
5.3c in
109
19
2.1c cm