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UNIT 6 : BASIC TRIGONOMETRY WITH TRIANGLES
LESSON
3: SINE LAW HOMEWORK QUESTIONS PAGE
2 (Solutions below)
Quick
Review: (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)
4.
Determine all possible solutions for each of the triangles below. Include a sketch.
Solutions for # 4:
D
660
11.1
E F
12.7
This is
the SSA case with < D 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 < D
since we know d.
R
9.5 7.1
h 7.1
540
P S Q
R
9.5 h 7.1
540
P r Q
This is
the SSA case with < P 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 (7.1) is opposite the given angle.
R
9.5 7.1
7.7 7.1
540
P Q
C
13.2 10.1 h
10.1
360
A D B
C
13.2 h 10.1
360
A c B
This is
the SSA case with < A 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 (10.1) is opposite the given angle.
C
13.2 10.1
360 129.80
A c D
D
14.1 11.4 h
11.4
43.20
E G F
D
14.1 h 11.4
43.20
E d F
This is
the SSA case with < E 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 (11.4) is opposite the given angle.
D
14.1 11.4
43.20 122.10
E d G
This is the AAS case.
There will be only one triangle.
D
470
f 8.4
1020
E d F
First
find < E: 1800 – (1020
+ 470) = 310
A
1020
c 7.9
B C
10.6
This is
the SSA case with < A obtuse; the side across from the given
angle is the longer side. There is one triangle only.
First we
must find a second angle. Find < B
since we know b.
P
1210
r 8.7
Q R
4.1
This is
the SSA case with < P obtuse ; the side across from the given
angle (1210) is the shorter side. There is no triangle possible.
P
1210
r 8.7
Q R
4.1