– To use the Law of Sines to solve triangles .

What You Will Learn: – To use the Law of Sines to solve triangles. – To find the area of triangles using trigonometry LawOfSines.ppt

sin A sin B sin C a b c = = Law of Sines- Let ABC be any triangle with a, b, and c representing the measures of sides opposite the angles with measures A, B, and C, respectively. Then Proof for the Law of Sines is algebraic. LawOfSines.ppt

sin B = sin A = b a h a h b h C D c sin A a sin B b = A B Given: ABC. b sin A = h = a sin B b sin A = a sin B LawOfSines.ppt

sin C = sin B = h c h b C sin C c sin C c sin B b sin A a sin B b = = A B Given: ABC. Reset the diagram b a h b sin C = h = c sin B b sin C = c sin B D c The Law of Sines = LawOfSines.ppt

NOTE: B 14.40 5 Adjacent side Opposite side 10o Given Angle A C There is a weakness with the Law of Sines. It can not directly determine, by calculation alone, if an unknown angle is obtuse. In certain situations it can have two solutions. If we are given SSA, the relationship of the two sides determine if there is NO solution, ONE solution, or TWO solutions to the triangle. The problem occurs when the side opposite the given angle is less than the given adjacent side to the given angle. LawOfSines.ppt

NOTE: B 14.40 5 10o C A For example: Suppose ABC has  A = 10o, side a = 5 cm and side c = 14.40cm. Solve the triangle. NOTE: from geometry we are given SSA, with the short side opposite the given angle. LawOfSines.ppt

=  sin–1(14.40  sin 10o / 5 )  30o sin 140o b B =  (5 sin 140o / sin 10o )  18.51 cm 14.40 Drawn to Scale 5 B A C B 14.40 14.40 18.51 5 5 5 150o sin C 14.40 sin 10o 5 sin 10o 5 10o A C C A C   B = 140o This happens when the side opposite the given angle is less than the second given side. However, given the original dimensions, there is another possible triangle. 9.85 LawOfSines.ppt

B 14.40 5 A 9.85 C So, the question is, “Which solution is correct?” They both are. You have to recognize that there are two solutions. The angle opposite the given adjacent side has a supplementary angle. In our example we found  C was 30o. It could also be 180o – 30o = 150o When you rework the problem you will find that Is also valid.  A = 10o,  B = 20o,  C = 150o LawOfSines.ppt

B 5 Drawn to Scale 14.40 C 25o A B Drawn to Scale 14.40 5 20.32o A C a b sin A  b 5  14.40 sin 25o  6.086 units (NO SOLUTION) Suppose  A = 20.32o, then a= b sin A 5 = 5.001 (ONE SOLUTION) LawOfSines.ppt

Possible solutions when given SSA parameters. ISBN 0-07-827999-2, page 727 LawOfSines.ppt

Y 14 33o 47o X Z sin100 y sin100 y sin33 x sin33 x sin33 x sin47 14 sin47 14 sin X x sin Y y sin Z z 14 sin 33 = x sin 47 = = = = 14 sin 100 = y sin 47 Definition: Solving the Triangle - Finding themeasuresof all theanglesandsidesof a triangle. 100o x 10.4 = y 18.9 14 sin 100 = y sin 47 14 sin 33 = x sin 47 Find m Y using the Third Angle Theorem m Y = 180o – 33o – 47o m Y = 100o Once we have solved for all the parts of a triangle, we have solve the triangle. 18.9 y 10.4 x LawOfSines.ppt

C b a A B c Area = ½ base height h = height = b sin A base = c Area = ½ cb sin A WLOG Area = ½ ac sin B Area = ½ ab sin C h The area of a triangle is one half the product of the lengths of two sides and the sine of their included angle. LawOfSines.ppt

sin C c sin A a sin B b = = What You Have Learned: – To use theLaw of Sinestosolve triangles. LawOfSines.ppt

END OF LINE LawOfSines.ppt

– To use the Law of Sines to solve triangles .