Geometric Trig SL1. The diagram below shows triangle PQR. The length of [PQ] is 7 cm, the length of [PR] is 10 cm, and is 75.

(a)Find (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 =

(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 = 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.

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, =110, = 52 and = 60.

(a)Find AD. (b)Find DE. (c)The area of triangle BCD is 5.68 cm2. Find .

(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)Calculate , 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.

(b)Find , giving your answer in radians.

(c)Given that is 1.63 radians, calculate the area of

(i)sector PAEB;(ii)sector OADB.

(d)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.

(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)(i)Explain why OPQ is an isosceles triangle.

(ii)Use the cosine rule to show that cos = .

(iii)Hence show that sin = .

(iv)Find the area of the triangle OPQ.

(b)Consider the smaller semi-circle, with centre P.

(i)Write down the size of (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 = .

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, BD = 25, =.

(a)Use the cosine rule to show that BD = .

Let = 40.

(b)(i)Find the value of sin .

(ii)Find the two possible values for the size of .

(iii)Given that is an acute angle, find the perimeter of ABCD.

(c)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 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].

(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 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 A1

sin R = 0.676148...

= 42.5A1N2

(b)P = 180 75 RP = 62.5(A1)

substitution into any correct formulaA1

eg area PQR = (their P)

= 31.0 (cm2)A1N2

2. (a)evidence of appropriate approachM1

eg 3 = r =13.5 (cm)A1N1

(b)adding two radii plus 3(M1)

perimeter = 27+3 (cm)(= 36.4)A1N2

(c)evidence of appropriate approachM1

eg area = 20.25 (cm2) (= 63.6)

3. (a)evidence of using area of a triangle(M1)

eg A = 2 sin A1N2

(b)METHOD 1 = (A1)

area OPA = (= 2 sin ( ))A1

since sin ( ) = sin R1

then both triangles have the same areaAGN0

METHOD 2triangle OPA has the same height and the same base as triangle OPBR3

then both triangles have the same areaAGN0

(c)area semi-circle = A1

area APB = 2 sin + 2 sin (= 4 sin )A1

S = area of semicircle area APB (= 2 4 sin )M1

S = 2( 2 sin )AGN0

(d)METHOD 1attempt to differentiate(M1)

eg setting derivative equal to 0(M1)

correct equationA1

eg 4 cos = 0, cos = 0, 4 cos = 0

= A

4. (a)Evidence of choosing cosine rule(M1)

eg a2 = b2 + c2 2bc cos ACorrect substitutionA1

eg (AD)2 = 7.12 + 9.22 2(7.1) (9.2) cos 60(AD)2 = 69.73(A1)

AD = 8.35 (cm)A1N2

(b)180 162 = 18(A1)

Evidence of choosing sine rule(M1)

Correct substitutionA1

eg =

DE = 2.75 (cm)A1N2

(c)Setting up equation(M1)

eg ab sin C = 5.68, bh = 5.68

Correct substitutionA1

eg 5.68 = (3.2) (7.1) sin , 3.2 h = 5.68, (h = 3.55)

sin = 0.5(A1)

30 and/or 150A1N2

(d)Finding AC (60 + DC)(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 )

eg (AC)2 = 9.22 + 3.22A1

AC = 9.74 (cm)A1N3

(e)For finding area of triangle ABD(M1)

Correct substitution Area = 9.2 7.1 sin 60A1

= 28.28...A1

Area of ABCD = 28.28... + 5.68(M1)

= 34.0 (cm2)A1N3

5. (a)METHOD 1Evidence of using the cosine rule(M1)

eg cos C = Correct substitution

eg cos = A1

cos = 0.25

= 1.82 (radians)A1N2

METHOD 2Area of AOBP = 5.81 (from part (d))

Area of triangle AOP = 2.905(M1)

2.9050 = 0.5 2 3 sinA1

= 1.32 or 1.82

= 1.82 (radians)A1N2

(b) = 2( 1.82)(= 2 3.64)(A1)

= 2.64 (radians)A1N2

(c)(i)Appropriate method of finding area(M1)

eg area = Area of sector PAEB = A1

= 13.0 (cm2)(accept the exact value 13.04)A1N2

(ii)Area of sector OADB = A1

= 11.9 (cm2)A1N1

(d)(i)Area AOBE = Area PAEB Area AOBP (= 13.0 5.81)M1

= 7.19 (accept 7.23 from the exact answer for PAEB)A1N1

(ii)Area shaded = Area OADB Area AOBE (= 11.9 7.19)M1

= 4.71 (accept answers between 4.63 and 4.72)A1N1

6. (a)For using perimeter = r + r + arc length(M1)

20 = 2r + rA1

AGN0

(b)Finding A = (A1)

For setting up equation in rM1

Correct simplified equation, or sketch

eg 10r r2 = 25, r2 10r + 25 = 0(A1)

r = 5 cmA1N2

7. (a)(i)OP = PQ (= 3cm)R1

So OPQ is isoscelesAGN0

(ii)Using cos rule correctly eg cos = (M1)

cos = A1

cos = AGN0

(iii)Evidence of using sin2 A + cos2 A = 1M1

sin = A1

sin = AGN0

(iv)Evidence of using area triangle OPQ = M1

eg Area triangle OPQ =

A1N1

(b)(i) = 1.4594...

= 1.46A1N1

(ii)Evidence of using formula for area of a sector(M1)

eg Area sector OPQ =

= 6.57A1N2

(c) = (A1)

Area sector QOS = A1

= 6.73A1N2

(d)Area of small semi-circle is 4.5 (= 14.137...)A1

Evidence of correct approachM1

eg Area = area of semi-circle area sector OPQ area sector QOS +area triangle POQ

Correct expressionA1

eg 4.5 6.5675... 6.7285... + 4.472..., 4.5 (6.7285... + 2.095...),

4.5 (6.5675... + 2.256...)

Area of the shaded region = 5.31A1N1

8. (a)METHOD 1Note:There are many valid algebraic approaches

to this problem (eg completing the square,

using . Use the following mark

allocation as a guide.(i)Using (M1)

32x + 160 = 0A1

x = 5A1N2

(ii)ymax = 16(52) + 160(5) 256

ymax = 144A1N1

METHOD 2(i)Sketch of the correct parabola (may be seen in part (ii))M1

x = 5

A2N2

(ii)ymax = 144A1N1

(b)(i)z = 10 x(accept x + z = 10)A1N1

(ii)z2 = x2 + 62 2 x 6 cos ZA2N2

(iii)Substituting for z into the expression in part (ii)(M1)

Expanding 100 20x + x2 = x2 + 36 12x cos ZA1

Simplifying 12x cos Z = 20x 64A1

Isolating cos Z = A1

cos Z = AGN0

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.(c)Evidence of using the formula for area of a triangle

M1

A1

A2 = 9x2 sin2 ZAGN0

(d)Using sin2 Z = 1 cos2 Z(A1)

Substituting for cos ZA1

for expanding A1

for simplifying to an expression that clearly leads to the required answerA1

eg A2 = 9x2 (25x2 160x + 256)

A2 = 16x2 + 160x 256AG

(e)(i)144 (is maximum value of A2, from part (a))A1

Amax = 12A1N1

(ii)IsoscelesA1N1

9. (a)For correct substitution into cosine ruleA1

BD = For factorizing 16, BD = A1

= AGN0

(b)(i)BD = 5.5653 ...(A1)

M1A1

sin = 0.911(accept 0.910, subject to AP)A1N2

(ii) = 65.7A1N1

Or = 180 their acute angle(M1)

= 114A1N2

(iii) = 89.3(A1)

(or cosine rule)M1A1

BC = 13.2(accept 13.17)A1

Perimeter = 4 + 8 + 12 + 13.2

= 37.2A1N2

(c)Area = 4 8 sin 40A1

= 10.3

10. (a)for using cosine rule (M1)

(A1) (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)

for using sine rule (may be implied)(M1)

(A1)

(A1)(N2)

(ii)

(A1)

(Accept) (A1)(N1)5

(c) from previous triangle Therefore alternative(A1)

(M1)(A1) (A1)(N1)4

(d)

Minimum length for BC when = 90or diagramshowing right triangle(M1)

(A1)(N1)2