Geometric Trig – SL
1. The diagram below shows triangle PQR. The length of [PQ] is 7 cm, the length of [PR] is 10 cm, and PQ̂R
is 75.
(a)
Find PQ̂R.
(b)
Find the area of triangle PQR.
2. The diagram below shows a circle centre O, with radius r. The length of arc ABC is 3 cm and AÔC =
(a)
Find the value of r.
(b)
Find the perimeter of sector OABC.
(c)
Find the area of sector OABC.
3. The following diagram shows a semicircle centre O,
diameter [AB], with radius 2.
Let P be a point on the circumference, with PÔB = 
radians.
(a)
Find the area of the triangle OPB, in terms of .
(b)
Explain why the area of triangle OPA is the
same as the area triangle OPB.
2π
.
9
Let S be the total area of the two segments shaded in
the diagram below.
(c)
Show that S = 2( − 2 sin  ).
(d)
Find the value of  when S is a local minimum.
(e)
Find a value of  for which S has its greatest
value.
4. The following diagram shows a pentagon ABCDE, with AB = 9.2 cm, BC = 3.2 cm, BD = 7.1 cm, AÊD
=110, AD̂E = 52 and AB̂D = 60.
(c)
The area of triangle BCD is 5.68 cm2. Find DB̂C .
(a)
Find AD.
(b)
Find DE.
(d)
Find AC.
(e)
Find the area of quadrilateral ABCD.
5. The following diagram shows the triangle AOP, where OP = 2 cm, AP = 4 cm and AO = 3 cm.
A
diagram not to
scale
O
(a)
P
Calculate AÔP , giving your answer in radians.
The following diagram shows two circles which intersect at the points A and B. The smaller circle C1 has
centre O and radius 3 cm, the larger circle C2 has centre P and radius 4 cm, and OP = 2 cm. The point D
lies on the circumference of C1 and E on the circumference of C2.Triangle AOP is the same as triangle
AOP in the diagram above.
A
C2
C1
O
E
D
diagram not to
scale
P
B
(b)
Find AÔB , giving your answer in radians.
(c)
Given that AP̂B is 1.63 radians, calculate the area of
(i)
(d)
sector PAEB;
(ii)
sector OADB.
The area of the quadrilateral AOBP is 5.81 cm2.
(i)
Find the area of AOBE.
(ii)
Hence find the area of the shaded region AEBD.
6. The following diagram shows a sector of a circle of radius r cm, and angle  at the centre. The perimeter of
the sector is 20 cm.
20  2r
.
r
(a)
Show that  =
(b)
The area of the sector is 25 cm2. Find the value of r.
7. The following diagram shows two semi-circles. The larger one has centre O and radius 4 cm. The smaller
one has centre P, radius 3 cm, and passes through O. The line (OP) meets the larger semi-circle at S. The
semi-circles intersect at Q.
(a)
(b)
(i)
Explain why OPQ is an isosceles triangle.
(ii)
Use the cosine rule to show that cos OP̂Q =
(iii)
Hence show that sin OP̂Q =
(iv)
Find the area of the triangle OPQ.
1
.
9
80
.
9
Consider the smaller semi-circle, with centre P.
(i)
Write down the size of OP̂Q.
(ii)
Calculate the area of the sector OPQ.
(c)
Consider the larger semi-circle, with centre O. Calculate the area of the sector QOS.
(d)
Hence calculate the area of the shaded region.
8. (a) Let y = –16x2 + 160x –256. Given that y has a maximum value, find
(i)
the value of x giving the maximum value of y;
(ii)
this maximum value of y.
The triangle XYZ has XZ = 6, YZ = x, XY = z as shown below. The perimeter of triangle XYZ is 16.
(b)
(i)
Express z in terms of x.
(ii)
Using the cosine rule, express z2 in terms of x and cos Z.
(iii)
Hence, show that cos Z =
5 x 16
.
3x
Let the area of triangle XYZ be A.
(c)
Show that A2 = 9x2 sin2 Z.
(d)
Hence, show that A2 = –16x2 + 160x – 256.
(e)
(i)
Hence, write down the maximum area for triangle XYZ.
(ii)
What type of triangle is the triangle with maximum area?
9. The diagram below shows a quadrilateral ABCD. AB = 4, AD = 8, CD =12, B Ĉ D = 25, BÂD =.
(a)
Use the cosine rule to show that BD = 4 5  4 cos  .
Let  = 40.
(b)
(c)
(i)
Find the value of sin CB̂D .
(ii)
Find the two possible values for the size of CB̂D .
(iii)
Given that CB̂D is an acute angle, find the perimeter of ABCD.
Find the area of triangle ABD.
10. The diagram shows a triangular region formed by a hedge [AB], a part of a river bank [AC] and a fence
[BC]. The hedge is 17 m long and BÂC is 29°. The end of the fence, point C, can be positioned
anywhere along the river bank.
(a)
Given that point C is 15 m from A, find the length of the fence [BC].
15 m
A
river bank
C
29°
17 m
B
(b)
The farmer has another, longer fence. It is possible for him to enclose two different triangular
regions with this fence. He places the fence so that AB̂C is 85°.
(i)
Find the distance from A to C.
(ii)
Find the area of the region ABC with the fence in this position.
(c)
To form the second region, he moves the fencing so that point C is closer to point A.
Find the new distance from A to C.
(d)
Find the minimum length of fence [BC] needed to enclose a triangular region ABC.
Solutions:
1. (a) choosing sine rule
(M1)
correct substitution
sin R sin 75

7
10
A1
sin R = 0.676148...
PR̂Q = 42.5
(b)
A1
P = 180  75  R
P = 62.5
(A1)
substitution into any correct formula
eg area  PQR =
A1
2. (a) evidence of appropriate approach
(b)
(c)
A1
1
 7 10  sin (their P)
2
= 31.0 (cm2)
eg 3 = r
N2
N2
M1
2
9
r =13.5 (cm)
A1
N1
adding two radii plus 3 (M1)
perimeter = 27+3 (cm) (= 36.4)
A1
N2
evidence of appropriate approach
eg
M1
1
2
13.5 2 
2
9
area = 20.25 (cm2) (= 63.6)
3. (a) evidence of using area of a triangle
(M1)
1
eg A   2  2  sin θ
2
A = 2 sin 
(b)
A1
N2
METHOD 1
PÔA =   
1
2  2  sin   θ  (= 2 sin (  ))
2
since sin (  ) = sin 
then both triangles have the same area
area OPA =
(A1)
A1
R1
AG
N0
R3
AG
N0
METHOD 2
triangle OPA has the same height and the same base as triangle OPB
then both triangles have the same area
(c)
(d)
1
2
 2  2 A1
2
area  APB = 2 sin  + 2 sin  (= 4 sin )
S = area of semicircle  area APB (= 2  4 sin )
S = 2( − 2 sin )
area semi-circle =
A1
M1
AG
N0
METHOD 1
attempt to differentiate
(M1)
ds
  4 cos θ
dθ
setting derivative equal to 0
(M1)
eg
correct equation
A1
eg 4 cos  = 0, cos  = 0, 4 cos  = 0
=

2
A
4. (a) Evidence of choosing cosine rule
(M1)
eg a2 = b2 + c2  2bc cos A
Correct substitution
A1
eg (AD)2 = 7.12 + 9.22  2(7.1) (9.2) cos 60
(AD)2 = 69.73
AD = 8.35 (cm)
(b)
A1
N2
180  162 = 18 (A1)
Evidence of choosing sine rule
Correct substitution
eg
(M1)
A1
8.35
DE
=
sin 110 
sin 18
DE = 2.75 (cm)
(c)
(A1)
A1
N2
Setting up equation (M1)
eg
1
1
ab sin C = 5.68,
bh = 5.68
2
2
Correct substitution
eg 5.68 =
A1
1
1
(3.2) (7.1) sin DB̂C ,  3.2  h = 5.68, (h = 3.55)
2
2
sin DB̂C = 0.5
DB̂C 30 and/or 150
(A1)
A1
N2
(d)
Finding A B̂ C (60 + D B̂ C)
(A1)
Using appropriate formula
(M1)
eg (AC)2 = (AB)2 + (BC)2, (AC)2 = (AB)2 + (BC)2  2 (AB)
(BC) cos ABC
Correct substitution (allow FT on their seen AB̂C )
(e)
eg (AC)2 = 9.22 + 3.22
A1
AC = 9.74 (cm)
A1
For finding area of triangle ABD
Correct substitution Area =
N3
(M1)
1
 9.2  7.1 sin 60
2
= 28.28...
A1
A1
Area of ABCD = 28.28... + 5.68
(M1)
= 34.0 (cm2)
A1
N3
5. (a) METHOD 1
Evidence of using the cosine rule
eg cos C =
(M1)
a2  b2  c2 2
, a  b 2  c 2  2bc cos A
2ab
Correct substitution
eg cos AÔP =
32  2 2  4 2 2 2
, 4  3  2 2  2  3 2 cos AÔP
2  3 2
A1
cos AÔP = 0.25
 26  
AÔP = 1.82  
 (radians)
 45 
A1
N2
METHOD 2
Area of AOBP = 5.81 (from part (d))
Area of triangle AOP = 2.905
(M1)
2.9050 = 0.5  2  3  sin AÔP
A1
AÔP = 1.32 or 1.82
 26  
 (radians)
 45 
AÔP = 1.82  
(b)
AÔB = 2(  1.82)
(= 2  3.64)
A1
(A1)
N2
 38 
= 2.64  
 (radians)
 45 
(c)
(i)
A1
Appropriate method of finding area
eg area =
(M1)
1 2
θr
2
Area of sector PAEB =
1 2
 4 1.63
2
A1
= 13.0 (cm2)
(accept the exact value 13.04)
(ii)
(d)
(i)
Area of sector OADB =
A1
1 2
 3  2.64
2
= 11.9 (cm2)
A1
Area AOBE = Area PAEB  Area AOBP (= 13.0  5.81)
M1
Area shaded = Area OADB  Area AOBE (= 11.9  7.19)
= 4.71 (accept answers between 4.63 and 4.72)
6. (a) For using perimeter = r + r + arc length
(b)
20  2r
r
AG

1 2  20  2r 
2
r 
 10r  r
2  r 

(a)
(i)
(ii)
(iii)
N1
N0
M1
OP = PQ (= 3cm)
So  OPQ is isosceles
Using cos rule correctly eg cos OP̂Q =
32  32  4 2
2  3 3
(A1)
A1
N2
R1
AG
N0
(M1)
cos OP̂Q =
9  9 16  2 
 
18
 18 
A1
cos OP̂Q =
1
9
AG
Evidence of using sin2 A + cos2 A = 1
N1
(A1)
For setting up equation in r
Correct simplified equation, or sketch
eg 10r – r2 = 25, r2 – 10r + 25 = 0
r = 5 cm
7.
A1
20 = 2r + r
Finding A =
N1
M1
(M1)
A1

N2
A1
= 7.19 (accept 7.23 from the exact answer for PAEB) A1
(ii)
N2
M1
N0
sin OP̂Q = 1 


  80 

81 

AG
1
 OP  PQ sin P
2
Evidence of using area triangle OPQ =
eg
(i)
80
2

20
  4.47 
Evidence of using formula for area of a sector
eg Area sector OPQ =
QÔP =
N1
A1
N1
1 2
 3 1.4594 
2
 1.4594 
 0.841
2
Area sector QOS =
A1
N2
(A1)
1 2
 4  0.841
2
A1
= 6.73
(d)
A1
(M1)
= 6.57
(c)
M1
OP̂Q = 1.4594...
OP̂Q = 1.46
(ii)
N0
1
80 9
3 3
,  0.9938 
2
9 2
Area triangle OPQ =
(b)
A1
80
9
sin OP̂Q =
(iv)
1
81
Area of small semi-circle is 4.5 (= 14.137...)
A1
N2
A1
Evidence of correct approach
eg Area = area of semi-circle  area sector OPQ  area sector QOS +
area triangle POQ
M1
Correct expression
eg 4.5  6.5675...  6.7285... + 4.472..., 4.5  (6.7285... + 2.095...),
A1
4.5 (6.5675... + 2.256...)
Area of the shaded region = 5.31
8. (a) METHOD 1
Note:
There are many valid algebraic approaches
to this problem (eg completing the square,
b
using x 
) . Use the following mark
2a
allocation as a guide.
A1
N1
(i)
(ii)
dy
0
dx
Using
(M1)
32x + 160 = 0
A1
x=5
A1
N2
A1
N1
ymax = 16(52) + 160(5)  256
ymax = 144
METHOD 2
(i)
(b)
Sketch of the correct parabola (may be seen in part (ii))
M1
x=5
A2
N2
(ii)
ymax = 144
A1
N1
(i)
z = 10  x
A1
N1
(ii)
z2 = x2 + 62 2  x  6  cos Z
A2
N2
(iii)
Substituting for z into the expression in part (ii)
(M1)
Expanding 100  20x + x2 = x2 + 36  12x cos Z
A1
Simplifying 12x cos Z = 20x  64
A1
(accept x + z = 10)
Isolating cos Z =
20 x  64
12 x
5 x 16
3x
Note: Expanding, simplifying and isolating may
be done in any order, with the final A1
being awarded for an expression that
clearly leads to the required answer.
cos Z =
(c)
(d)
A1
AG
N0
Evidence of using the formula for area of a triangle
1


 A   6  x  sin Z  M1
2


1


A  3x sin Z  A 2   3 6 x 2  sin 2 Z 
4


A1
A2 = 9x2 sin2 Z
AG
Using sin2 Z = 1  cos2 Z
Substituting
(A1)
5 x 16
for cos Z
3x
2
 25 x 2 160 x  256 
 5 x 16 

for expanding 
 to 

9x 2
 3x 


A1
A1
for simplifying to an expression that clearly leads to the required answer A1
N0
eg A2 = 9x2  (25x2  160x + 256)
A2 = 16x2 + 160x  256
AG
144 (is maximum value of A2, from part (a))
A1
Amax = 12
A1
N1
Isosceles
A1
N1
9. (a) For correct substitution into cosine rule
A1
(e)
(i)
(ii)
4 2  8 2  2  4  8 cos θ
BD =
For factorizing 16, BD = 165  4 cos θ 
= 4 5  4 cos θ
(b)
(i)
BD = 5.5653 ...
(accept 0.910, subject to AP)
CB̂D = 65.7
Or CB̂D = 180  their acute angle
= 114
A1
N2
A1
N1
(M1)
A1
BD̂C = 89.3
N2
(A1)
BC
5.5653
BC
12
(or cosine rule)

or

sin 89.3 sin 25 sin 89.3 sin 65.7
BC = 13.2
N0
M1A1
sin CB̂D = 0.911
(iii)
AG
(A1)
sin CB̂D sin 25

12
5.5653
(ii)
A1
(accept 13.17…)
M1A1
A1
Perimeter = 4 + 8 + 12 + 13.2
= 37.2
(c)
Area =
A1
N2
1
 4  8  sin 40 A1
2
= 10.3
10. (a)
for using cosine rule  a 2  b2  c 2  2ab cos C 
(M1)
BC2  152  17 2  2 15 17  cos 29
(A1)
BC  8.24 m
(A1) (N0)
3
Notes: Either the first or the second line may be implied, but not both.
Award no marks if 8.24 is obtained by assuming a right (angled)
triangle (BC = 17 sin 29).
(i)
C
A
29°
17
85°
B
A CB  180  (29  85)  66
for using sine rule (may be implied)
(M1)
AC
17

sin85 sin 66
AC 
(A1)
17sin85
sin 66
AC  (18.5380 )  18.5 m
(ii)
Area 
(A1) (N2)
1
17 18.538... sin 29
2
(A1)
 76.4 m2 (Accept 76.2 m2 )
(c)
(A1) (N1)
5
AĈB from previous triangle  66
ˆ  180  66  114 (or 29  85)
Therefore alternative ACB
(A1)
ˆ  180  (29  114)  37
ABC
C
114°
A
29°
17
37°
B
AC
17

sin37 sin114
AC  (11.19906 )  11.2 m
(M1)(A1)
(A1) (N1)
4
(d)
A
C
29°
17
B
Minimum length for BC when AĈB = 90°or diagram
showing right triangle
sin 29 
(M1)
CB
17
CB  17sin 29
CB  (8.2417 )  8.24 m
(A1) (N1)
2