Law of Sines In Section 14-3 you studied techniques for solving - PowerPoint PPT Presentation

Section 14-4 Law of Sines

In Section 14-3 you studied techniques for solving right triangles. In this section you will learn techniques for solving non-right triangles or oblique triangles .

C a b A c B

To solve an oblique triangle, you need to know the measure of at least one side and any two other parts of the triangle.

In this section we will solve oblique triangles knowing the following Two angles and one side (ASA or AAS)

Law of Sines If ABC is a triangle with sides , , a b and , then c sin sin sin A B C   . a b c

Example 1 Find the remaining angle and sides. C 102.3 ° 27.4 ft a 28.7 ° A c B

To find A subtract B and C from 180 ° . A = 180 – B – C A = 180 – 28.7 ° – 102.3 ° A = 49.0 °

sin A sin B sin C   a b c Substitute 27.4 for b solve for a . sin49 sin28.7  a 27.4 27.4sin49   43.06 ft a sin28.7

sin A sin B sin C   a b c Substitute 27.4 for b solve for c . sin102.3 sin28.7  c 27.4 27.4sin102.3   55.75 ft c sin28.7

Example 2 Find the remaining angle and sides. C b 32 ft 30 ° 45 ° A c B

To find C subtract A and B from 180 ° . C = 180 – A – B C = 180 – 30 ° – 45 ° C = 105 °

sin A sin B sin C   a b c Substitute 32 for a solve for b . sin45 sin30  b 32 32sin45   45.25 ft b sin30

sin A sin B sin C   a b c Substitute 32 for a solve for c . sin105 sin30  c 32 32sin105   61.82 ft c sin30

Example 3

A surveyor locates points A and B at the same elevation and 3950 ft. apart. At A , the angle of elevation to the summit of the mountain is 18 ° . At B , the angle of elevation is 31 ° . (a) Find the distance from B to the summit. (b) Find the height of the mountain.

C x 18 ° 31 ° A B 3950 ft. To find x we need to find the m  ACB . We must first find the m  ABC . m  ABC = 180 ° − 31° = 149 °

C x 18 ° 149 ° 31 ° A B 3950 ft. m  ACB = 180 ° − 18° − 149° = 13 ° Now we use the law of sines to solve for x .

C 13 ° x 18 ° 31 ° A B 3950 ft.   sin13 sin18  3950 x  3950sin18   5426 ft. x  sin13

C 13 ° h 5426 18 ° 31 ° A B 3950 ft. Now use the trig ratios from Section 14-3 to the height.   5426sin31 h h   sin31  2795 ft. 5426

Area of an Oblique Triangle The area of any triangle is one-half the product of the lengths of two sides and the sine of the included angle. That is,

C 1  Area bc sin A 2 a b 1  Area sin ac B 2 B c A 1  Area ab sin C 2

Example 4 Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102 ° .

Let a = 90 meters 52 meters 102 ° b = 52 meters 90 meters C = 102 ° 1      Area 90 52 sin102 2  2 Area 2288.87 m