6. No Steps Solutions
1 Draw the line UP and
shade the plane RPQ U
S
P
R
Q
T
Question 1
The Diagram shows a triangular prism. Identify the angle
between the line UP and the plane RPQ
7. No Steps Solutions
2. 1. Arrange the line and plane
in two rows
2. Find out the same alphabet – point P
(identify the point of line UP that touches the plane RPQ, )
3. Draw 3 boxes,
4. Write point P in the middle and point U in
the first box (point does not touch the plane)
5. Look at U, choose Which One is the Nearest
to U (WON) (Non-slashed alphabets)
Between point R and Q, point R will be
chosen and write it in the third box.
6. Angle between line UP to the plane RPQ
is
U P
R P Q
U P
R P Q
U P R
U
S
P
R
Q
T
ÐUPR
8. No Steps Solutions
1 Draw the line QT and
shade the plane PQRS
T U
P
V
S
R
W
8 cm
Q
16 cm
6 cm
Question 2
The Diagram shows a cuboid with a horizontal rectangular
base PQRS. Calculate the angle between the line QT and
the base PQRS
9. No Steps Solutions
2. 1. Arrange the line and plane
in two rows
2. Find out the same alphabet – point Q
(identify the point of line QT that touches the plane PQRS, )
3. Draw 3 boxes,
4. Write point Q in the middle and point T in
the first box. (point does not touch the plane)
5. Look at T, choose Which One is the
Nearest to T (WON) (Non-slashed alphabets-
Between point P, R or S, point S will be
chosen and write it in the third box.
6. Angle between line UP to the plane RPQ
is
Q T
P Q R S
Q T
P Q R S
T Q S
T U
ÐTQS P
V
S
W
Q
R
16 cm
6 cm
8 cm
10. No Steps Solutions
3 Refer to the points you
have obtained in step 2.
(point T,Q,S) Draw ΔTQS.
Mark the right angle.
P
V
S
W
Q
U
R
T
16 cm
6 cm
8 cm
T
S Q
11. No Steps Solutions
4 With the information
given in the question,
label the length of the
sides of ΔTQS. At least
two sides of the length
must be known. Use
Pythagoras Theorem if
necessary
P
V
S
W
Q
U
R
T
16 cm
6 cm
8 cm
T
S Q
8 cm
162 +62 =17.09
12. No Steps Solutions
5 Mark
the opposite side,TS
the adjacent side,SQ
P
V
S
W
Q
U
R
T
16 cm
6 cm
8 cm
T
S Q
8 cm
162 +62 =17.09
13. ÐTQS
tanÐTQS = 8
ÐTQS = 25.08°or25°5'
P
V
S
W
Q
U
R
16 cm
6 cm
8 cm
T
17.09
T
S Q
8 cm
are 162 +62 =17.09
14.
15. No Steps Solutions
1 Shade the plane PBC the
plane BCRQ
9 cm
S
P Q
U
T
R
12 cm
C
D
15 cm
A B
Question 4
The diagram shows a prism with cross section BCRQ.
Given T and U are the midpoint of AD and BC respectively,
P and Q are right above T and U respectively and PQRS
is a square.
16. No Steps Solutions
2. 1. Arrange the plane PCB and plane
BCRQ in two rows
2. Find out the same alphabet
(identify the points of PCB that touch the plane BCRQ, )
3. Draw 3 boxes,
4. Write point CB in the middle and point P in
the first box.
P C B
B C R Q
P C B
B C R Q
P CB
U
T
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
17. No Steps Solutions
2. 5. Look at P, choose WON – Slashed
alphabet
BU = CU, SO CHOOSE THE MIDPOINT OF
BC (U)
6. Look at P choose Which One is the
Nearest to P (WON) (Non-slashed
alphabets - point at the plane BCRQ,
point Q is the nearest to P, Write point
Q in the third box.
7. Angle between plane PCB to the plane
BCRQ is
P C B
B C R Q
P BC
P U Q
U
T
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
ÐPUQ
18. No Steps Solutions
3 Refer to the points you
have obtained in step 2.
(point P,U,Q) Draw
ΔPUQ. Mark the right
angle.
P
Q
U
U
T
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
19. No Steps Solutions
4 With the information
given in the question,
label the length of the
sides of ΔPUQ. At least
two sides of the length
must be known. Use
Pythagoras Theorem if
necessary
P
Q
U
U
T
9 cm
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
12 cm
20. No Steps Solutions
5 Mark
the opposite side PQ
the adjasent side UQ
P
Q
U
U
T
9 cm
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
12 cm
21. ÐPUQ
Q
tanÐPUQ = 9
12
ÐPUQ = 36.87°or36°52'
P
U
9 cm
12 cm
U
T
C
9 cm
S
D
A B
R
P Q
15 cm
12 cm
are
22. No Steps Solutions
1 Shade the plane PRV and
the plane QRVU
4 cm
T R
U
R
Q
S
P
5 cm
W V
12 cm
Question 5
The diagram shows a cuboid with base TUVW, Calculate the
angle between PRV and the plane QRVU
23. No Steps Solutions
2. 1. Arrange the plane PRV and plane
QRVU in two rows
2. Find out the same alphabet
(identify the points of PRV that touch the plane QRVU, )
3. Draw 3 boxes,
4. Write point RV at the middle and point P in
the first box.
P R V
Q R V U
P R V
Q R V U
P RV
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
R
24. No Steps Solutions
2. 5. Look at P, choose WON – Slashed
alphabet (Between point R and V, point
which is nearer to point P. Point R is chosen
and write it in the middle box.
6. Look at P choose Which One is the
Nearest to P (WON) (Non-slashed
alphabets – Between point Q and and U,
point Q is chosen and write it in the third
box,
7. Angle between plane PRV to the plane
QRUV is
P R V
Q R V U
P RV
P R Q
ÐPRQ
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
R
25. No Steps Solutions
3 Refer to the points you
have obtained in step 2.
(point P,R,Q) Draw
ΔPRQ. Mark the right
angle.
P
Q
R
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
R
26. No Steps Solutions
4 With the information
given in the question,
label the length of the
sides of ΔPRQ. At least
two sides of the length
must be known. Use
Pythagoras Theorem if
necessary
P
Q
R
12 cm
5 cm
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
Diagram 5
R
27. No Steps Solutions
5 Mark
the opposite side,PQ
the adjacent side, RQ
P
Q
R
12 cm
5 cm
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
R
28. ÐPRQ
Q
tanÐPRQ = 12
5
ÐPRQ = 67.38°or67°22'
P
U
12 cm
5 cm
R
Q
S
P
T U
5 cm
W V
12 cm
4 cm
R
are
29. Try this..
H
E
U G
T
L M
N
R
F
P
Name the angle between the plane LUM
with the plane LPNM
Answer : URT
30. Find the angle between the plane JFE
and the plane DEF.
L
F D
E
J
M
5
5
13
5
