Solve the right Solve the right triangle.triangle.
sin 378
8sin 37
4.815
a
a
a
cos378
8cos37
6.389
b
b
b
90 37 53 B
Solve the right Solve the right triangle.triangle.
1
12sin
1312
sin13
67.380
A
A
A
1
12cos
1312
cos13
22.620
B
B
B
5a
Oblique TrianglesOblique Triangles
But what if the triangle isn’t right? We need a method that will work for what we call oblique triangles.
(Triangles that aren’t right)
Law of SinesLaw of Sines
AA
Usually used when Usually used when you have ASA or you have ASA or AAS AAS
In In ANYANY triangle ABC: triangle ABC:
(Notice that each (Notice that each angle goes with its angle goes with its corresponding side corresponding side in each proportion. in each proportion. One of the One of the proportions must proportions must have a “known” have a “known” angle and side)angle and side)
B C
c b
a
sin A sin B sin C
a b c
Example: Find the Example: Find the missing variablemissing variable In the triangle below, mIn the triangle below, mA = 33A = 33oo, ,
mmB = 47B = 47oo & b = 14, find a. & b = 14, find a.
A B
Csin 47 sin 33
14 asin 47 14sin 33
10.426
a
a
4733
14
You know both B’s, the angle and the side, so that will be our “known”
ExampleExample
A civil engineer A civil engineer wants to determine wants to determine the distances from the distances from points A and B to an points A and B to an inaccessible point C, inaccessible point C, as shown. From as shown. From direct measurement, direct measurement, the engineer knows the engineer knows that AB = 25m, that AB = 25m, A = 110 A = 110oo, and , and B = 20B = 20oo. Find . Find AC and BC. AC and BC.
A
B
C
ExampleExample
1.1. Mark the Mark the drawing with drawing with known info known info
2.2. C = 180 – (110 + 20)C = 180 – (110 + 20)
= 50= 50oo
33..
4.4.
A
B
C
25110110oo
2020oo
sin 110 sin 50
a 25
25sin110 sin 50
30.7
a
a
sin 20 sin 50
b 25
25sin 20 sin 50
11.2
b
b
You DoYou Do
Find ALL missing angles and sides in the Find ALL missing angles and sides in the triangle below: mtriangle below: mA = 28A = 28oo, a = 12 & b , a = 12 & b =24=24
A
B
CmmB = 70B = 70oo
mmC = 82C = 82oo
C = 25.312C = 25.312
Ambiguous CaseAmbiguous Case
When two sides and a non-included angle are given, there are several situations possible – this is called the ambiguous case.
There could be only one triangle.
There could be two triangles (when an angle(s) has more than one possibility).
There could be no triangle (when you take the inverse sine of a value larger than 1).
ExampleExample
Find angle B in triangle ABC if a = Find angle B in triangle ABC if a = 2, b = 6, and A = 302, b = 6, and A = 30oo
Applying the Law of Sines, we Applying the Law of Sines, we have :have :osin30 sinB
2 66sin30 2sinB
6sin30sinB
2 1.5 sinB
Since sinB can never be larger than 1, this triangle does not exist.
Example 2Example 2
Find the missing Find the missing parts in triangle ABC parts in triangle ABC if a = 54cm, b = if a = 54cm, b = 62cm, and A = 4062cm, and A = 40oo
First solve for BFirst solve for B Since B can be 48Since B can be 48oo
or 132or 132oo, C can be , C can be 9292oo or 8 or 8oo
Then c can be 84cm Then c can be 84cm or 12cm (using Law or 12cm (using Law of Sines)of Sines)
o
1 o o
sin40 sinB54 62
62sin40 54sinB
62sin40sinB
54 .7380 sinB
sin B 48 or 132
Thus, there are 2 possibilities and no way to tell which is correct—ambiguous!
Example 3-ApplicationExample 3-Application
A forest ranger at an observation point (A) sights a fire in the direction 32° east of north. Another ranger at a second observation point (B), 10 miles due east of A, sights the same fire 48° west of north. Find the distance from each observation point to the fire.
10
32o
48o
58o 42o
A B
80o 8.6116.795