|
|
|
|
|
|||||||
|
|
|||||||||||
UNIT 6 : BASIC TRIGONOMETRY WITH TRIANGLES
LESSON
6: UNIT TEST
Test Questions: (Solutions below)
1. Find PQ in the diagram below.
P
T
18.2 8.6 cm
105.1o
Q R 12.3
cm
S
3. Determine the side “a” in each of the
following. Decide whether there are two
solutions, one solution or no solution.
4. Solve each of the following triangles
5. From the top of a 83 m cliff, the angles of
depression of ships A and B at sea are 170 and 280
respectively. Find the distance (AB)
between them.
6. A surveyor wishes to find the height of a
hill AD in the diagram. The angle of elevation of the top of the hill from
point B is 13.60. A base
line BC of length
150 m is
staked out and angles DBC and DCB are shown in the diagram. Find the height h of the hill.
7. Nomar travels due north for 3.5 h at 80
km/h. From there he travels N500E
at 100 km/h for 2.5h. How far is he
from his starting point.
(Solutions)
1. Find PQ in the diagram below.
P
T
18.2 cm
8.6 cm
105.1o
Q R 12.3 cm S
Solution: We must use the triangle with 3 elements given.
Solution:
x = -3 y = 4 r = 5
3. Determine the side “a” in each of the
following. Decide whether there are two
solutions, one solution or no solution.
A
6.3 h 3.0
320
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
6.6 4.1
h 4.1
320
B D a
E 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 (4.1) is opposite the given angle.
A
6.6 4.1
320 121.50
B a D
A
7.9 12.3
B 67.50 C
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.
4. Solve each of the following triangles
A
98.50
16.7
B C
20.3
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.
11.4
P R
13.2 8.9
Q
Solution:
This is the SSS case. Use the cosine law for any angle, say angle
P.
5. From the top of a 83 m cliff, the angles of
depression of ships A and B at sea are 170 and 280
respectively. Find the distance (AB)
between them.
Solution:
6. A surveyor wishes to find the height of a
mountain AD in the diagram. The angle of elevation of the top of the mountain
from point B is 13.60. A
base line BC of length 150 m is staked out and angles DBC and DCB are shown in
the diagram. Find the height h
of the mountain.
Solution:
Again
use the triangle with 3 elements known.
This would be triangle DBC.
This is the AAS case. Use the sine law.
150
m
7. Nomar travels due north for 3.5 h at 80
km/h. From there he travels N500E
at 100 km/h for 2.5h. How far is he
from his starting point.
Solution:
80 km/h
x 3.5 h = 280 km due north.[OP]
100 km/h
x 2.5 h = 250 km N500E [PQ]
