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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 HOMEWORK QUESTIONS PAGE 1 (Solutions below)

Homework questions: (Solutions below)

C

b h b a

A₁ ( B

D A₂

Homework Questions: (all measurements in cm. unless otherwise stated. Give answers to the nearest tenth)

1. In each of the following triangles, the side opposite the given angle is the shorter side. Find angles m and n in each case to the nearest tenth of a degree..

a)

C

8 h 8 12

A m n 32⁰ B

D E

b)

2. In each of the following, determine which triangles have no solution, one solution or two solutions.

a)

A

7.5 10.8

B 41⁰

b)

13.4

P R

12.3

71⁰

Q

c)

A

3.2 6.3

B 32⁰ C

d) 12.4

D F

46⁰

8.5

E

e)

M 32⁰ P

16.4 m

75⁰

N

f) 12.4

D F

8.5

115⁰

E

3. Determine the side “a” in each of the following. Decide whether there are two solutions, one solution or no solution.

4. Determine all possible solutions for each of the triangles below. Include a sketch.

Solutions for # 1-3:

1. In each of the following triangles, the side opposite the given angle is the shorter side. Find angles m and n in each case to the nearest tenth of a degree..

a)

C

8 h 8 12

A m n 32⁰ B

D E

Solution:

This is the SSA case with a given acute angle.

First we calculate the height of the triangle since the shorter side is opposite the given angle.

C

8 12

A m 32⁰ B

C

8 12

E 127.4⁰ 32⁰ B

C

8 h 8 12

A 52.6⁰ 127.4⁰ 32⁰ B

D E

b)

Solution:

This is the SSA case with a given acute angle.

First we calculate the height of the triangle since the shorter side is opposite the given angle.

2. In each of the following, determine which triangles have no solution, one solution or two solutions.

a)

A

7.5 h 10.8

B 41⁰ C

D

Solution:

This is the SSA case with the side opposite the given acute angle being the shorter side.

First we calculate the height of the triangle.

A

7.5 h 7.5 10.8

B 41⁰ C

D E

b)

13.4

P R

12.3

71⁰

Q

Solution:

Since the side opposite the given acute angle (71⁰) is the longer side, we have the SSA case with one solution.

c)

A

c =3.2 6.3

h=3.3

B 32⁰ C

D

Solution:

This is the SSA case with the side opposite the given acute angle being the shorter side.

First we calculate the height of the triangle.

d) 12.4

D F

46⁰

8.5

E h 8.5

H

G

Solution:

This is the SSA case with the side opposite the given acute angle being the shorter side.

First we redraw the diagram to show both possible triangles (DEF and DGF) and the height FH

Next we calculate the height of the triangle.

12.4

D F

46⁰

8.5

E h 8.5

H

G

e)

M 32⁰ P

16.4 m

75⁰

N

Solution:

This is the AAS case, two angles and a side – there is one solution.

f) 12.4

D F

8.5

115⁰

E

Solution:

This is the SSA case with an obtuse angle given. The side opposite the given angle is the larger side.

Hence there is one solution.

3. Determine the side “a” in each of the following. Decide whether there are two solutions, one solution or no solution.

Solutions:

A

3.1 h 1.5

34⁰

B a C

This is the SSA case with < B acute; the side across from the given angle is the shorter side. This is the ambiguous case.

First we calculate the height of the triangle since the shorter side (1.5) is opposite the given angle.

A

13.2 8.1 h 8.1

31⁰

B D a C

This is the SSA case with < B acute; the side across from the given angle is the shorter side. This is the ambiguous case

First we calculate the height of the triangle since the shorter side (8.1) is opposite the given angle.

A

13.2 8.1

31⁰ 122.9⁰

B a D

A

8.1 11.4

B 68⁰ C

a

This is the SSA case with < B acute; the side across from the given angle is the longer side. There is one triangle only.

First we must find a second angle. Find < C since we know c.

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