jdlogo

jdlogo

jdlogo

jdlogo

jdlogo

Home

Right Triangles

Angles in Standard Position

Sine Law & Ambiguous Case

Cosine Law

Problem Solving

Summary&Test

 

jdsmathnotes

 

 


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]

 

 

 

 

 

 

 

 

 

 

 

 

Return to top of page

Click here to return to unit review