Warm up   A 5.2 m ladder leans against a wall. The

bottom of the ladder is 1.9 m from the wall. What angle does the ladder make with the ground (to the nearest degree)?

Cos-1(1.9/5.2) = 69o

Objective: To use the Law of Sines in order to solve oblique triangles

The Law of Sines

Consider the first category, an acute triangle (, , are acute).

hNow, sin( ) , so that h a sin( )

ah

But sin( ) , so that h c sin( )c

By transitivity, a sin( ) c sin( )

sin( ) sin( )Which means

c a

Create an altitude, h.

Theorem Law of Sines

The Law of Sines is used when we know any two angles and one side or when we know two sides and an angle opposite one of those sides.

Applying the Law of Sines The Law of Sines may be used when the

known parts of the triangle are:◦ 1. one side and two angles (SAA), (ASA)◦ 2. two sides and an angle opposite one of the

sides (SSA)

Example: In triangle ABC, angle A = 106 o, angle B =

31o and side a = 10 cm. Solve the triangle ABC by finding angle C and sides b and c.(round answers to 1 decimal place).

Use the fact that the sum of all three angles of a triangle is equal to 180 o to write an equation in C. A + B + C = 180 o Solve for C. C = 180 o - (A + B) = 43 o

Solution

Use sine law to write an equation in b. a / sin(A) = b / sin(B) Solve for b. b = a sin (B) / sin(A) = (approximately) 5.4 cm

Use the sine law to write an equation in c.a / sin(A) = c / sin(C)Solve for c. c = a sin (C) / sin(A) = (approximately) 7.1 cm

Solution (cont’d)

Area of an Oblique Triangle

The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle.

Referring to the triangles below, that each triangle has a height of h = b sin A.

A is acute. A is obtuse.

Area of a Triangle - SASSAS – you know two sides: b, c and

the angle between: A

Remember area of a triangle is ½ base ● height

Base = bHeight = c ● sin A Area K= ½ bc(sinA)

A

B

C

c a

b

h

Looking at this from all three sides:K = ½ ab(sin C) = ½ ac(sin B) = ½ bc (sin A)

Example: Find the area ofgiven a = 32 m, b = 9 m, and

ABC36 .m C

132 9 sin36

2Area m m

284.6Area m

You can also find the area if you know one side and 2 angles based on the Law of Sines.

so, substitute for b in the last

equation, K = ½ bc(sinA) gives you

Area of a Triangle

C

c

BSin

b

sin

C

Bcb

sin

sin

C

BAcK

sin

sinsin

2

1 2

Find the area of triangle JKL if j=45.7, K=111.1o, and L=27.3o.

Area of a Triangle

http://www.emathematics.net/trigonometria.php?tr=5

Law of Sines practice

Using the triangle above, A = 50o, B = 65o and a = 12.  Solve the triangle.

Warm up

Lesson 5-7 Law of Sines the Ambiguous Case

Objective: To determine whether a triangle has zero, one or two solutions and solve

using the Law of Sines.

The Ambiguous Case – SSA

In this case, you may have information that results in one triangle, two triangles, or no triangles.

SSA – No Solution

Two sides and an angle opposite one of the sides.

By the law of sines,

sin(57 ) sin( )

15 20

Thus,20 sin(57 )

sin( )15

20 (0.8387)sin( )

15sin( ) 1.1183 Impossible!

Therefore, there is no value for that exists! No Solution!

SSA – Two Solutions

By the law of sines,

sin(32 ) sin( )

30 42

So that,

42 sin(32 )sin( )

3042 (0.5299)

sin( )30

sin( ) 0.7419

48 or 132

Case 1 Case 2

48 32 180

100

132 32 180

16

Both triangles are valid! Therefore, we have two solutions.

Case 1 )sin(100 sin(32 )

c 30

30 sin(100 )c

sin(32 )

30 (0.9848)c

0.5299

c 55.7539

Case 2 sin(16 ) sin(32 )

c 30

30 sin(16 )c

sin(32 )

30 (0.2756)c

0.5299

c 15.6029

For SSA Triangles:

1. If A < 90°a. a < b

1. a < b(sin A) No Solution2. a = b(sin A) 1 Solution3. a > b(sin A) 2 Solution

b. a ≥ b 1 Solution

2. If A ≥ 90°a. a ≤ b No Solution

b. a > b 1 Solution

Practice Solve the triangle: A = 42°, a = 11, and b =

6