# IBDP Maths AI: Topic SL 3.3: Construction of labelled diagrams: IB style Questions SL Paper 1

## Question

On a map three schools A, B and C are situated as shown in the diagram.

Schools A and B are 625 metres apart.

Angle ABC = 102° and BC = 986 metres. Find the distance between A and C.

a.

Find the size of angle BAC.

b.

## Markscheme

Unit penalty (UP) is applicable in question part (a) only.

$${\text{AC}}^2 = 625^2 + 986^2 – 2 \times 625 \times 986 \times \cos102^\circ$$     (M1)(A1)

( = 1619072.159)

AC = 1272.43

(UP) =1270 m     (A1)     (C3)[3 marks]

a.

$$\frac{{986}}{{\operatorname{sinA} }} = \frac{{1270}}{{\sin 102^\circ }}$$     (M1)(A1)(ft)

$${\text{A}} = 49.4^\circ$$     (A1)(ft)

OR

$$\frac{{986}}{{\operatorname{sinA} }} = \frac{{1272.43}}{{\sin 102^\circ }}$$     (M1)(A1)(ft)

$${\text{A}} = 49.3^\circ$$     (A1)(ft)

OR

$$\cos {\text{A}} = \left( {\frac{{{{625}^2} + {{1270}^2} – {{986}^2}}}{{2 \times 625 \times 1270}}} \right)$$     (M1)(A1)(ft)

$${\text{A}} = 49.5^\circ$$     (A1)(ft)     (C3)[3 marks]

b.

## Question

Triangle $${\text{ABC}}$$ is such that $${\text{AC}}$$ is $$7{\text{ cm}}$$, angle $${\text{ABC}}$$ is $${65^ \circ }$$ and angle $${\text{ACB}}$$ is $${30^ \circ }$$.

Sketch the triangle writing in the side length and angles.

a.

Calculate the length of $${\text{AB}}$$.

b.

Find the area of triangle $${\text{ABC}}$$.

c.

## Markscheme (A1)     (C1)

Note: (A1) for fully labelled sketch.
[1 mark]

a.

Unit penalty (UP) may apply in this question.

$$\frac{{{\text{AB}}}}{{\sin 30}} = \frac{7}{{\sin 65}}$$     (M1)

(UP)     $${\text{AB}} = 3.86{\text{ cm}}$$     (A1)(ft)     (C2)

Note: (M1) for use of sine rule with correct values substituted.[2 marks]

b.

Unit penalty (UP) may apply in this question.

$${\text{Angle BAC}} = {85^ \circ }$$     (A1)

$${\text{Area}} = \frac{1}{2} \times 7 \times 3.86 \times \sin {85^ \circ }$$     (M1)

(UP)     $$= 13.5{\text{ }}{{\text{cm}}^2}$$     (A1)(ft)     (C3)[3 marks]

c.

## Question

The diagram shows triangle ABC in which angle BAC $$= 30^\circ$$, BC $$= 6.7$$ cm and AC $$= 13.4$$ cm. Calculate the size of angle ACB.

a.

Nadia makes an accurate drawing of triangle ABC. She measures angle BAC and finds it to be 29°.

Calculate the percentage error in Nadia’s measurement of angle BAC.

b.

## Markscheme

$$\frac{{\sin {\text{A}}{\operatorname{\hat B}}{\text{C}}}}{{13.4}} = \frac{{\sin 30^\circ }}{{6.7}}$$     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitution.

$${\text{A}}{\operatorname{\hat B}}{\text{C}}$$ = 90°     (A1)

$${\text{A}}{\operatorname{\hat C}}{\text{B}}$$ = 60°     (A1)(ft)     (C4)

Note: Radians give no solution, award maximum (M1)(A1)(A0).[4 marks]

a.

$$\frac{{29 – 30}}{{30}} \times 100$$     (M1)

Note: Award (M1) for correct substitution into correct formula.

% error = −33.3 %     (A1)     (C2)

Notes: Percentage symbol not required. Accept positive answer.[2 marks]

b.

## Question

The diagram shows a triangle ABC in which AC = 17 cm. M is the midpoint of AC.
Triangle ABM is equilateral. Write down the size of angle MCB.

a.1.

Write down the length of BM in cm.

a.i.

Write down the size of angle BMC.

a.ii.

Calculate the length of BC in cm.

b.

## Markscheme

30°     (A1)     (C3)[1 mark]

a.1.

8.5 (cm)     (A1)[1 mark]

a.i.

120°     (A1)[1 mark]

a.ii.

$$\frac{{{\text{BC}}}}{{\sin 120}} = \frac{{8.5}}{{\sin 30}}$$     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

$${\text{BC}} = {\text{14}}{\text{.7}}\left( {\frac{{17\sqrt 3 }}{2}} \right)$$     (A1)(ft)[3 marks]

b.

## Question

The base of a prism is a regular hexagon. The centre of the hexagon is O and the length of OA is 15 cm. Write down the size of angle AOB.

a.

Find the area of the triangle AOB.

b.

The height of the prism is 20 cm.

Find the volume of the prism.

c.

## Markscheme

60°     (A1)     (C1)[1 mark]

a.

$$\frac{{15 \times \sqrt {{{15}^2} – {{7.5}^2}} }}{2} = 97.4{\text{ c}}{{\text{m}}^2}$$     (97.5 cm2)     (A1)(M1)(A1)

Notes: Award (A1) for correct height, (M1) for substitution in the area formula, (A1) for correct answer.

Accept 97.5 cm2 from taking the height to be 13 cm.

OR

$$\frac{1}{2} \times {15^2} \times \sin 60^\circ = 97.4{\text{ c}}{{\text{m}}^2}$$     (M1)(A1)(A1)(ft)     (C3)

Notes: Award (M1) for substituted formula of the area of a triangle, (A1) for correct substitution, (A1)(ft) for answer.

If radians used award at most (M1)(A1)(A0).[3 marks]

b.

97.4 × 120 = 11700 cm3     (M1)(A1)(ft)     (C2)

Notes: Award (M1) for multiplying their part (b) by 120.[2 marks]

c.

## Question

In the diagram, $${\text{AD}} = 4{\text{ m}}$$, $${\text{AB}} = 9{\text{ m}}$$, $${\text{BC}} = 10{\text{ m}}$$, $${\text{B}}\hat {\text{D}}{\text{A}} = {90^ \circ }$$ and $${\text{D}}\hat {\text{B}}{\text{C}} = {100^ \circ }$$ . Calculate the size of $${\text{A}}\hat {\text{B}}{\text{C}}$$.

a.

Calculate the length of AC.

b.

## Markscheme

$$\sin {\text{A}}\hat {\text{B}}{\text{D}} = \frac{4}{9}$$     (M1)

$$100 + {\text{their }}({\text{A}}\hat {\text{B}}{\text{D}})$$     (M1)

$$126\%$$     (A1)     (C3)

Notes: Accept an equivalent trigonometrical equation involving angle ABD for the first (M1).

Radians used gives $$100\%$$ . Award at most (M1)(M1)(A0) if working shown.

$${\text{BD}} = 8{\text{ m}}$$ leading to $$127\%$$ . Award at most (M1)(M1)(A0) (premature rounding).[3 marks]

a.

$${\text{A}}{{\text{C}}^2} = {10^2} + {9^2} – 2 \times 10 \times 9 \times \cos (126.38 \ldots )$$     (M1)(A1)

Notes: Award (M1) for substituted cosine formula. Award (A1) for correct substitution using their answer to part (a).

$${\text{AC}} = 17.0{\text{ m}}$$     (A1)(ft)     (C3)

Notes: Accept $$16.9{\text{ m}}$$ for using $$126$$. Follow through from their answer to part (a). Radians used gives $$5.08$$. Award at most (M1)(A1)(A0)(ft) if working shown.[3 marks]

b.

## Question

The diagram shows quadrilateral ABCD in which AB = 13 m , AD = 6 m and DC = 10 m. Angle ADC =120° and angle ABC = 40°. Calculate the length of AC.

a.

Calculate the size of angle ACB.

b.

## Markscheme

AC2 = 62 + 102 – 2×10×6×cos120°     (M1)(A1)

Note: Award (M1) for substitution in cosine formula, (A1) for correct substitutions.

AC = 14 (m)     (A1)     (C3)[3 marks]

a.

$$\frac{{14}}{{\sin 40}} = \frac{{13}}{{\operatorname{sinACB} }}$$     (M1)(A1)(ft)

Note: Award (M1) for substitution in sine formula, (A1) for correct substitutions.

Angle ACB = 36.6° (36.6463…)     (A1)(ft)     (C3)

Note: Follow through from their (a).[3 marks]

b.

## Question

In triangle ABC, BC = 8 m, angle ACB = 110°, angle CAB = 40°, and angle ABC = 30°. Find the length of AC.

a.

Find the area of triangle ABC.

b.

## Markscheme

$$\frac{{{\text{AC}}}}{{\sin 30^\circ }} = \frac{8}{{\sin 40^\circ }}$$     (M1)(A1)

Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.

AC = 6.22 (m) (6.22289…)     (A1)     (C3)[3 marks]

a.

Area of triangle $${\text{ABC}} = \frac{1}{2} \times 8 \times 6.22289… \times \sin 110^\circ$$     (M1)(A1)(ft)

Note: Award (M1) for substitution in the correct formula, (A1)(ft) for their correct substitutions. Follow through from their part (a).

Area triangle ABC = 23.4 m2 (23.3904…m2)     (A1)(ft)     (C3)

Note: Follow through from a positive answer to their part (a). The answer is 23.4 m2, units are required.[3 marks]

b.

## Question

In the diagram, triangle ABC is isosceles. AB = AC and angle ACB is 32°. The length of side AC is x cm. Write down the size of angle CBA.

a.

Write down the size of angle CAB.

b.

The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.

c.

## Markscheme

32°     (A1)     (C1)[1 mark]

a.

116°     (A1)     (C1)[1 mark]

b.

$$360 = \frac{1}{2} \times {x^2} \times \sin 116^\circ$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into correct formula with 360 seen, (A1)(ft) for correct substitution, follow through from their answer to part (b).

x = 28.3 (cm)     (A1)(ft)

x = 283 (mm)     (A1)(ft)     (C4)

Notes: The final (A1)(ft) is for their cm answer converted to mm. If their incorrect cm answer is seen the final (A1)(ft) can be awarded for correct conversion to mm.[4 marks]

c.

## Question

The quadrilateral ABCD has AB = 10 cm, AD = 12 cm and CD = 7 cm.

The size of angle ABC is 100° and the size of angle ACB is 50°. Find the length of AC in centimetres.

a.

Find the size of angle ADC.

b.

## Markscheme

$$\frac{{{\text{AC}}}}{{\sin 100^\circ }} = \frac{{10}}{{\sin 50^\circ }}$$     (M1)(A1)

Note: Award (M1) for substitution in the sine rule formula, (A1) for correct substitutions.

$$=12.9 (12.8557…)$$     (A1)     (C3)

a.

$$\frac{{{{12}^2} + {7^2} – {{12.8557…}^2}}}{{2 \times 12 \times 7}}$$     (M1)(A1)(ft)

Note: Award (M1) for substitution in the cosine rule formula, (A1)(ft) for correct substitutions.

= 80.5° (80.4994…°)     (A1)(ft)     (C3)

Notes: Follow through from their answer to part (a). Accept 80.9° for using 12.9. Using the radian answer from part (a) leads to an impossible triangle, award (M1)(A1)(ft)(A0).

b.

## Question

The diagram shows a triangle $${\rm{ABC}}$$. The size of angle $${\rm{C\hat AB}}$$ is $$55^\circ$$ and the length of $${\rm{AM}}$$ is $$10$$ m, where $${\rm{M}}$$ is the midpoint of $${\rm{AB}}$$. Triangle $${\rm{CMB}}$$ is isosceles with $${\text{CM}} = {\text{MB}}$$. Write down the length of $${\rm{MB}}$$.

a.

Find the size of angle $${\rm{C\hat MB}}$$.

b.

Find the length of $${\rm{CB}}$$.

c.

## Markscheme

$$10$$ m     (A1)(C1)

a.

$${\rm{A\hat MC}} = 70^\circ \;\;\;$$OR$$\;\;\;{\rm{A\hat CM}} = 55^\circ$$     (A1)

$${\rm{C\hat MB}} = 110^\circ$$     (A1)     (C2)

b.

$${\text{C}}{{\text{B}}^2} = {10^2} + {10^2} – 2 \times 10 \times 10 \times \cos 110^\circ$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the cosine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

$$\frac{{{\text{CB}}}}{{\sin 110^\circ }} = \frac{{10}}{{\sin 35^\circ }}$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the sine rule formula, (A1)(ft) for correct substitution. Follow through from their answer to part (b).

OR

$${\rm{A\hat CB}} = 90^\circ$$     (A1)

$$\sin 55^\circ = \frac{{{\text{CB}}}}{{55}}\;\;\;$$OR$$\;\;\;\cos 35^\circ = \frac{{{\text{CB}}}}{{20}}$$     (M1)

Note: Award (A1) for some indication that $${\rm{A\hat CB}} = 90^\circ$$, (M1) for correct trigonometric equation.

OR

Perpendicular $${\rm{MN}}$$ is drawn from $${\rm{M}}$$ to $${\rm{CB}}$$.     (A1)

$$\frac{{\frac{1}{2}{\text{CB}}}}{{10}} = \cos 35^\circ$$     (M1)

Note: Award (A1) for some indication of the perpendicular bisector of $${\rm{BC}}$$, (M1) for correct trigonometric equation.

$${\text{CB}} = 16.4{\text{ (m)}}\;\;\;\left( {16.3830 \ldots {\text{ (m)}}} \right)$$     (A1)(ft)(C3)

Notes: Where a candidate uses $${\rm{C\hat MB}} = 90^\circ$$ and finds $${\text{CB}} = 14.1{\text{ (m)}}$$ award, at most, (M1)(A1)(A0).

Where a candidate uses $${\rm{C\hat MB}} = 60^\circ$$ and finds $${\text{CB}} = 10{\text{ (m)}}$$ award, at most, (M1)(A1)(A0).

c.

## Question

In the following diagram, ABCD is the square base of a right pyramid with vertex V. The centre of the base is O. The diagonal of the base, AC, is 8 cm long. The sloping edges are 10 cm long. Write down the length of $${\text{AO}}$$.

a.

Find the size of the angle that the sloping edge $${\text{VA}}$$ makes with the base of the pyramid.

b.

Hence, or otherwise, find the area of the triangle $${\text{CAV}}$$.

c.

## Markscheme

$${\text{AO}} = 4{\text{ (cm)}}$$     (A1)     (C1)

a.

$$\cos {\rm{O\hat AV}} = \frac{4}{{10}}$$     (M1)

Note: Award (M1) for their correct trigonometric ratio.

OR

$$\cos {\rm{O\hat AV}} = \frac{{{{10}^2} + {8^2} – {{10}^2}}}{{2 \times 10 \times 8}}\;\;\;$$OR$$\;\;\;\frac{{{{10}^2} + {4^2} – {{(9.16515 \ldots )}^2}}}{{2 \times 10 \times 4}}$$     (M1)

Note: Award (M1) for correct substitution into the cosine rule formula.

$${\rm{O\hat AV}} = 66.4^\circ \;\;\;(66.4218 \ldots )$$     (A1)(ft)     (C2)

b.

$${\text{area}} = \frac{{8 \times 10 \times \sin (66.4218 \ldots ^\circ)}}{2}\;\;\;$$OR$$\;\;\;\frac{1}{2} \times 8 \times \sqrt {{{10}^2} – {4^{\text{2}}}}$$

OR$$\;\;\;\frac{1}{2} \times 10 \times 10 \times \sin (47.1563 \ldots ^\circ )$$     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitutions. Follow through from their answer to part (b) and/or part (a).

$${\text{area}} = 36.7{\text{ c}}{{\text{m}}^2}\;\;\;(36.6606 \ldots {\text{ c}}{{\text{m}}^2})$$     (A1)(ft)     (C3)

Notes: Accept an answer of $$8\sqrt {21} {\text{ c}}{{\text{m}}^2}$$ which is the exact answer.

c.

## Question

When Bermuda $${\text{(B)}}$$, Puerto Rico $${\text{(P)}}$$, and Miami $${\text{(M)}}$$ are joined on a map using straight lines, a triangle is formed. This triangle is known as the Bermuda triangle.

According to the map, the distance $${\text{MB}}$$ is $$1650\,{\text{km}}$$, the distance $${\text{MP}}$$ is $$1500\,{\text{km}}$$ and angle $${\text{BMP}}$$ is $$57^\circ$$. Calculate the distance from Bermuda to Puerto Rico, $${\text{BP}}$$.

a.

Calculate the area of the Bermuda triangle.

b.

## Markscheme

$${\text{B}}{{\text{P}}^2} = {1650^2} + {1500^2} – 2 \times 1650 \times 1500\,\cos \,(57^\circ )$$                  (M1)(A1)

$$1510\,({\text{km}})\,\,\,\left( {1508.81…\,({\text{km}})} \right)$$       (A1)     (C3)

Notes: Award (M1) for substitution in the cosine rule formula, (A1) for correct substitution.

a.

$$\frac{1}{2} \times 1650 \times 1500 \times \sin \,57^\circ$$        (M1)(A1)

$$= 1\,040\,000\,({\text{k}}{{\text{m}}^2})\,\,\,\left( {1\,037\,854.82…\,({\text{k}}{{\text{m}}^2})} \right)$$       (A1)    (C3)

Note: Award (M1) for substitution in the area of triangle formula, (A1) for correct substitution.

b.

## Question

A triangular postage stamp, ABC, is shown in the diagram below, such that $${\text{AB}} = 5{\text{ cm}},{\rm{ B\hat AC}} = 34^\circ ,{\rm{ A\hat BC}} = 26^\circ$$ and $${\rm{A\hat CB}} = 120^\circ$$. Find the length of BC.

a.

Find the area of the postage stamp.

b.

## Markscheme

$$\frac{{{\text{BC}}}}{{\sin 34^\circ }} = \frac{5}{{\sin 120^\circ }}$$     (M1)(A1)

Note:     Award (M1) for substituted sine rule formula, (A1) for correct substitutions.

$${\text{BC}} = 3.23{\text{ (cm) }}\left( {3.22850 \ldots {\text{ (cm)}}} \right)$$     (A1)     (C3)[3 marks]

a.

$$\frac{1}{2}(5)(3.22850)\sin 26^\circ$$     (M1)(A1)(ft)

Note:     Award (M1) for substituted area of a triangle formula, (A1) for correct substitutions.

$$= 3.54{\text{ }}({\text{c}}{{\text{m}}^2}){\text{ }}\left( {3.53820 \ldots {\text{ }}({\text{c}}{{\text{m}}^2})} \right)$$     (A1)(ft)     (C3)

Note:     Follow through from part (a).[3 marks]

b.

## Question

Emily’s kite ABCD is hanging in a tree. The plane ABCDE is vertical.

Emily stands at point E at some distance from the tree, such that EAD is a straight line and angle BED = 7°. Emily knows BD = 1.2 metres and angle BDA = 53°, as shown in the diagram T is a point at the base of the tree. ET is a horizontal line. The angle of elevation of A from E is 41°.

Find the length of EB.

a.

Write down the angle of elevation of B from E.

b.

Find the vertical height of B above the ground.

c.

## Markscheme

Units are required in parts (a) and (c).

$$\frac{{{\text{EB}}}}{{\sin 53{\rm{^\circ }}}} = \frac{{1.2}}{{\sin 7{\rm{^\circ }}}}$$     (M1)(A1)

Note:     Award (M1) for substitution into sine formula, (A1) for correct substitution.

$$({\text{EB}} = ){\text{ }}7.86{\text{ m}}$$$$\,\,\,$$OR$$\,\,\,$$$$786{\text{ cm }}(7.86385 \ldots {\text{ m}}$$$$\,\,\,$$OR$$\,\,\,$$$$786.385 \ldots {\text{ cm}})$$     (A1)     (C3)[3 marks]

a.

34°     (A1)     (C1)[1 mark]

b.

Units are required in parts (a) and (c).

$$\sin 34^\circ = \frac{{{\text{height}}}}{{7.86385 \ldots }}$$     (M1)

Note:     Award (M1) for correct substitution into a trigonometric ratio.

$$({\text{height}} = ){\text{ }}4.40{\text{ m}}$$$$\,\,\,$$OR$$\,\,\,$$$$440{\text{ cm }}(4.39741 \ldots {\text{ m}}$$$$\,\,\,$$OR$$\,\,\,$$$$439.741 \ldots {\text{ cm}})$$     (A1)(ft)     (C2)

Note:     Accept “BT” used for height. Follow through from parts (a) and (b). Use of 7.86 gives an answer of 4.39525….[2 marks]

c.

## Question

Two fixed points, A and B, are 40 m apart on horizontal ground. Two straight ropes, AP and BP, are attached to the same point, P, on the base of a hot air balloon which is vertically above the line AB. The length of BP is 30 m and angle BAP is 48°. Angle APB is acute.

On the diagram, draw and label with an x the angle of depression of B from P.

a.

Find the size of angle APB.

b.

Find the size of the angle of depression of B from P.

c.

## Markscheme (A1) (C1)[1 mark]

a.

$$\frac{{40}}{{{\text{sin APB}}}} = \frac{{30}}{{{\text{sin 48}}^\circ }}$$     (M1)(A1)

Note: Award (M1) for substitution into sine rule, (A1) for correct substitution.

(angle APB =) 82.2°  (82.2473…°)     (A1)  (C3)[3 marks]

b.

180 − 48 − 82.2473…     (M1)

49.8°  (49.7526…°)     (A1)(ft)  (C2)

Note: Follow through from parts (a) and (b).[2 marks]

c.

## Question

A park in the form of a triangle, ABC, is shown in the following diagram. AB is 79 km and BC is 62 km. Angle A$$\mathop {\text{B}}\limits^ \wedge$$C is 52°. Calculate the length of side AC in km.

a.

Calculate the area of the park.

b.

## Markscheme

(AC2 =) 622 + 792 − 2 × 62 × 79 × cos(52°)     (M1)(A1)

Note: Award (M1) for substituting in the cosine rule formula, (A1) for correct substitution.

63.7  (63.6708…) (km)     (A1) (C3)[3 marks]

a.

$$\frac{1}{2}$$ × 62 × 79 × sin(52°)     (M1)(A1)

Note: Award (M1) for substituting in the area of triangle formula, (A1) for correct substitution.

1930 km2  (1929.83…km2)     (A1) (C3)[3 marks]

b.