     Home Right Triangles Angles in Standard Position Sine Law & Ambiguous Case Cosine Law Problem Solving Summary&Test UNIT 6  : BASIC TRIGONOMETRY WITH TRIANGLES

LESSON 3:  SINE LAW

The Sine Law:

The sine law is used to solve oblique triangles, that is triangles which are not right angled.

A   c                                             b B                             a                                           C

Given two angles and a side (AAS case)

Example 1:   A

c                                               b B      48o                                                               29o        C

22.4 cm

Solution:

First find the measure of  < A:    180o – (48o + 29o) = 103o

Now find c using the sine law: Now find b using the sine law: Given two sides and an ACUTE angle (< 900) opposite one of the sides [the non-contained angle] (SSA)

Example 2: Note:  Here the longer side (23.1) is opposite the given acute angle < E is acute.  There is only one triangle in this case.  D

16.4 cm                                                23.1 cm E      56.4o                                                                         F

Solution: Now find < D:   180 – (56.40 + 36.30) = 87.30

Next find d using the sin law: Given two sides and an ACUTE angle where the shorter side (b) is opposite the given angle (< B) (SSA)

Example 3:

Solution:  There are 3 possibilities here, one of which is the so-called ambiguous case where there are two triangles which satisfy the given conditions.  C

b                                               a A                                                                               (      B    C

b                 h                            a  A                                                                              (        B

D     C

b                 h            b              a  A1                                                                             (        B

D                        A2      C

9.6                h            9.6           12.4  A1                                                                          420      B

A2

Hence there are two triangles to solve here.

Triangle 1:  C

9.6                                              12.4 A1                                                                         420        B Triangle 2 : Recall from above that < CA2B = 120.20  C

9.6                            12.4 A2  120.20         420       B  C b=h                            a  A                                                      B     C

b                 h            b              a

A                                         A  (        B

The SSA case where the given angle B is OBTUSE (> 900)

Example 4:   C

b                     a A                                       B   C

b                     a A                                       B

Example 4(a):   C

13.5                   10.4  A                                       B

1010

Solution: This is the SSA case with a given obtuse angle and the side opposite this angle is the longer side.

There will be only one triangle to solve.

First find < A using the sine law.  