Revision - Trigonometry IB10
1.
The figure shows two adjacent triangular fields ABC and ACD where AD = 30 m, CD = 80 m,
BC = 50m. A D̂ C = 60° and B Â C = 30°.
D
80 m
60°
C
30 m
50 m
Note:
Diagram not
drawn to scale
30°
B
A
(a)
Using triangle ACD calculate the length AC.
(b)
Calculate the size of A B̂ C.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(Total 8 marks)
Revision IB10
April 2013
M. Wærn
2.
The figure below shows a hexagon with sides all of length 4 cm and with centre at O. The
interior angles of the hexagon are all equal.
F
A
O
E
B
4 cm
C
D
The interior angles of a polygon with n equal sides and n equal angles (regular polygon) add up
to (n – 2) × 180°.
(a)
Calculate the size of angle A B̂ C.
(b)
Given that OB = OC, find the area of the triangle OBC.
(c)
Find the area of the whole hexagon.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 8 marks)
Revision IB10
April 2013
M. Wærn
3.
The diagram shows a circle of radius R and centre O. A triangle AOB is drawn inside the circle.
The vertices of the triangle are at the centre, O, and at two points A and B on the circumference.
Angle AÔB is 110 degrees.
A
B
R
110°
R
O
(a)
Given that the area of the circle is 36 cm2, calculate the length of the radius R.
(b)
Calculate the length AB.
(c)
Write down the side length L of a square which has the same area as the given circle.
Working:
Answers:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 6 marks)
Revision IB10
April 2013
M. Wærn
4.
An old tower (BT) leans at 10 away from the vertical (represented by line TG).
The base of the tower is at B so that MB̂T = 100 .
Leonardo stands at L on flat ground 120 m away from B in the direction of the lean.
He measures the angle between the ground and the top of the tower T to be BL̂T = 26.5 .
T
not to scale
M
(a)
MB = 200
90°
100°
26.5°
B G
BL = 120
L
(i)
Find the value of angle BT̂ L .
(ii)
Use triangle BTL to calculate the sloping distance BT from the base, B to the top,
T of the tower.
(5)
(b)
Calculate the vertical height TG of the top of the tower.
(2)
(c)
Leonardo now walks to point M, a distance 200 m from B on the opposite side of the
tower. Calculate the distance from M to the top of the tower at T.
(3)
(Total 10 marks)
5.
Amir needs to construct an isosceles triangle ABC whose area is 100 cm2.
The equal sides, AB and BC, are 20 cm long.
(a)
Angle AB̂C is acute. Show that the angle AB̂C is 30 .
(2)
(b)
Find the length of AC.
(3)
(Total 5 marks)
Revision IB10
April 2013
M. Wærn
6.
Sylvia is making a square-based pyramid. Each triangle has a base of length 12 cm and a height
of 10 cm.
O
(not to scale)
M
(a)
Show that the height of the pyramid is 8 cm.
(2)
M is the midpoint of the base of one of the triangles and O is the apex of the pyramid.
(b)
Find the angle that the line MO makes with the base of the pyramid.
(3)
(c)
Calculate the volume of the pyramid.
(2)
(d)
Daniel wants to make a rectangular prism with the same volume as that of Sylvia’s
pyramid. The base of his prism is to be a square of side 10 cm. Calculate the height of the
prism.
(2)
(Total 9 marks)
Revision IB10
April 2013
M. Wærn
Answers Revision - Trigonometry IB10
1.
(a)
AC2 = CD2 + AD2 – 2 × AD × CD × cos (CD̂A)
2
2
= 80 + 30 – 2 × 30 × 80 × cos (60°)
AC2 = 4900
so AC = 70 m (units not required)
(b)
50
sin (30 )
70
(M1)
(A1)
(A1)
(A1) (C4)
(M1)
sin (AB̂C)
1 70 7
sin (AB̂C)
2 50 10
7
AB̂C sin –1
= 44.4°
10
(M1)(A1)
(A1) (C4)
Note: Accept 44° 25 ± 1 .
[8]
2.
(b)
(a)
Sum of internal angles is (n – 2) × 180 = (6 – 2) × 180 = 720°
720
Hence AB̂C
= 120°
6
OB̂C = 60°
(A1) (C3)
(M1)
so AOBC =
(c)
(M1)(A1)
1
× 4 × 2 × tan (60°) = 6.93 cm2 (units not required)
2
Total area of hexagon is 6 × AOBC
= 6 × 6.93 = 41.6 cm2 (units not required)
(A1)(A1) (C3)
(M1)
(A1) (C2)
[8]
R2 = 36 so R = 6 cm
3.
(a)
(b)
Use cosine rule. AB2 = 62 + 62 – 2
AB2 = 96.6
(M1)(A1) (C2)
6
6 cos (110 )
(M1)(A1)(ft)
AB = 9.83 cm
(A1)(ft)
OR
6
sin (35 )
AB
sin (110 )
(M1)(A1)(ft)
AB = 9.83 cm
(A1)(ft)
OR
1
AB
sin (55 ) = 2
6
110
55
2
(M1)(A1)(ft)
AB = 9.83
(A1)(ft) (C3)
1
AB must be evident to obtain the (M1) and the
2
first (A1) requires the 55 and the 6 to be correct.
Note: If this method is used, then the
(c)
L=
36
or 6
or 10.6 cm
(A1) (C1)
[6]
4.
(a)
(i)
Angle BT̂ L = 180 – 80 – 26.5 or 180 – 90 – 26.5 + 10
Revision IB10
April 2013
M. Wærn
(M1)
= 73.5
(ii)
(b)
(c)
BT
sin (26.5 )
(A1)(G2)
120
sin (73.5 )
BT = 55.8 m (3s.f.)
(M1)(A1)(ft)
(A1)(ft) (UP)
5
TG = 55.8 sin (80 ) or 55.8 cos (10 )
= 55.0 m (3s.f.)
Note: Apply (AP) if 0 missing
(M1)
(A1)(ft)(G2)
2
MT2 = 2002 + 55.82 – 2
(M1)(A1)(ft)
200
55.8
cos (100 )
MT = 217 m (3s.f.)
(A1)(ft)
(UP)
3
Notes: Follow through only from part (a)(ii).
Award marks at discretion for any valid alternative method. If radian mode has been
used throughout the question, award (A0) to the first incorrect answer then follow
through, but negative lengths are always awarded (A0)(ft). The answers are (all 3s.f.)
(a)(ii) 124 m (A0)(ft) (b) 123 m (A1)(ft)
(c) 313 m (A1)(ft)
[10]
5.
Unit penalty (UP) is applicable where indicated.
(a)
1 2
20 sin B 100
2
(M1)(A1)
B = 30
Note: (M1) for correct substituted formula and (A1) for correct substitution.
B=30 must be seen or previous (A1) mark is lost.
(b)
2
AC = 2
202− 2
202
cos 30
(AG)
2
(M1)(A1)
AC = 10.4 cm
(A1)(G2) (UP)
Note: (M1) for using cosine rule, (A1) for correct substitution. Last (A1) is for the correct answer.
Accept use of sine rule or any correct method e.g. AC = 2
20 sin 15 .
[5]
6.
Unit penalty (UP) is applicable where indicated.
(a)
x2 + 62 = 102
(A1)(M1)
x = 8 cm
Notes: (A1) for 6 (or 36) seen and (M1) for using Pythagoras with correct substitution.
x = 8 must be seen or previous (M1) mark is lost.
(b)
cos
=
6
10
vol =
(A1)(G2)
12 2 8
3
3
(M1)
= 384 cm3
(d) Let h be the height
2
(M1)(A1)
= 53.1
(c)
(AG)
(A1)(G2) (UP)
102 h = 384
2
(M1)
h = 3.84 cm
(A1)(ft)(G2)
(UP)
2
[9]
Revision IB10
April 2013
M. Wærn