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.[3]

a.

Find the size of angle BAC.[3]

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.[1]

a.

Calculate the length of $${\text{AB}}$$.[2]

b.

Find the area of triangle $${\text{ABC}}$$.[3]

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.[4]

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.[2]

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.[1]

a.1.

Write down the length of BM in cm.[1]

a.i.

Write down the size of angle BMC.[1]

a.ii.

Calculate the length of BC in cm.[3]

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.[1]

a.

Find the area of the triangle AOB.[3]

b.

The height of the prism is 20 cm.

Find the volume of the prism.[2]

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 $${\text{A}}\hat {\text{B}}{\text{C}}$$.[3]

a.

Calculate the length of AC.[3]

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.[3]

a.

Calculate the size of angle ACB.[3]

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.[3]

a.

Find the area of triangle ABC.[3]

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.[1]

a.

Write down the size of angle CAB.[1]

b.

The area of triangle ABC is 360 cm2. Calculate the length of side AC. Express your answer in millimetres.[4]

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.[3]

a.

Find the size of angle ADC.[3]

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 $${\rm{MB}}$$.[1]

a.

Find the size of angle $${\rm{C\hat MB}}$$.[2]

b.

Find the length of $${\rm{CB}}$$.[3]

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 $${\text{AO}}$$.[1]

a.

Find the size of the angle that the sloping edge $${\text{VA}}$$ makes with the base of the pyramid.[2]

b.

Hence, or otherwise, find the area of the triangle $${\text{CAV}}$$.[3]

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, $${\text{BP}}$$.[3]

a.

Calculate the area of the Bermuda triangle.[3]

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.[3]

a.

Find the area of the postage stamp.[3]

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.[3]

a.

Write down the angle of elevation of B from E.[1]

b.

Find the vertical height of B above the ground.[2]

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.[1]

a.

Find the size of angle APB.[3]

b.

Find the size of the angle of depression of B from P.[2]

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.[3]

a.

Calculate the area of the park.[3]

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.