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Date November 2019 Marks available 5 Reference code 19N.2.SL.TZ0.T_5
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Determine and Justify Question number T_5 Adapted from N/A

Question

Haraya owns two triangular plots of land, ABC and ACD. The length of AB is 30m, BC is 50m and AC is 70m. The size of DA^C is 55° and AD^C is 72°.

The following diagram shows this information.

Haraya attaches a 20m long rope to a vertical pole at point B.

Find the length of AD.

[4]
a.

Find the size of AB^C.

[3]
b.

Calculate the area of the triangular plot of land ABC.

[3]
c.

Determine whether the rope can extend into the triangular plot of land, ACD. Justify your answer.

[5]
d.

Markscheme

AC^D=53° (or equivalent)       (A1)

Note: Award (A1) for 53° (or equivalent) seen.

ADsin53°=70sin72°       (M1)(A1)

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

OR

AD2= 60.29152+702-2×70×60.2915×cos53       (A1)(M1)(A1)

Note: Award (A1) for 53 or 60.2915... seen, (M1) for substitution into cosine rule formula, (A1) for correct substitution.

AD= 58.8 m 58.7814       (A1)(G3)

[4 marks]

a.

cosAB^C=302+502-7022×30×50       (M1)(A1)

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

AB^C= 120°       (A1)(G2)

[3 marks]

b.

Units are required in part (c)

 

A=12×50×30×sin120°       (M1)(A1)(ft)

Note: Award (M1) for substitution into the area formula, (A1)(ft) for correct substitution. Award (M0)(A0)(A0) for 12×50×30.

A= 650m2  649.519m2       (A1)(ft)(G2)

Note: Follow through from part (b).

[3 marks]

c.

METHOD 1 (equating part (c) to expression for area of triangle ABC)

649.519=12×70×h       (M1)(A1)(ft)

Note: Award (M1) for correctly substituted area of triangle formula. Award (A1)(ft) for equating the area formula to their area found in part (c).

h= 18.6m 18.5576       (A1)(ft)

Note: Follow through from their part (c).

20>18.5576       (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle ABC”.

the rope passes inside the triangular plot of land ACD       (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

 

METHOD 2 (finding CA^B or AC^B with sine rule and then trig ratio)

sinCA^B50=sin120°50 CA^B=38.2132°       (M1)

Note: Award (M1) for their correct substitution into sine rule formula to find CA^B or AC^B. Follow through from their part (b).

h= 30×sin38.2132°       (M1)

Note: Award (M1) for correct substitution of their CA^B or AC^B into trig formula.

h= 18.6m 18.5576       (A1)(ft)

Note: Follow through from their part (b).

20>18.5576       (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle ABC”.

the rope passes inside the triangular plot of land ACD       (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

 

METHOD 3 (finding CA^B or AC^B with with cosine rule and then trig ratio)

cosAC^B=502+702-30225070 AC^B=21.7867°       (M1)

Note: Award (M1) for for their correct substitution into cosine rule formula to find CA^B or AC^B.

h= 50×sin21.7867°       (M1)

Note: Award (M1) for correct substitution of their CA^B or AC^B into trig formula.

h= 18.6m 18.5576       (A1)(ft)

20>18.5576       (R1)(ft)

Note: Accept “the length of the rope is greater than the altitude of triangle ABC”.

the rope passes inside the triangular plot of land ACD       (A1)(ft)

Note: Follow through from their altitude. The final (A1) is contingent on (R1) being awarded.

 

METHOD 4 (finding area of triangle with height 20, justifying the contradiction)

A=127020=700m2       (M1)(A1)

Note: Award (M1) for correct substitution into area of a triangle formula for a triangle with height 20 and base 70. Award (A1) for 700. Award (M0)(A0) for unsupported 700 unless subsequent reasoning explains how the 700 was found.

700>649.519       (R1)

if rope exactly touches the AC then this triangle has an area greater than
ABC and as the distance AC is fixed the altitude must be less than 20              (R1)

OR

127020>1270 (height perpendicular to AC) and therefore 20>height perpendicular to AC       (R1)(ft)

Note: Award (R1) for an explanation that recognizes the actual triangle ABC and this new triangle have the same base 70 and hence the height of triangle ABC is less than 20.

therefore, the rope passes inside the triangular plot of land ACD       (A1)(ft)

 

Note: Other methods, besides those listed here, may be possible. These methods can be summarized in two broad groups: the first is to find the altitude of the triangle, and compare it to 20, and the second is to create an artificial triangle with an altitude of 20 and explain why this triangle is not ABC by relating to area and the given lengths of the sides.

 

[5 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 3—Geometry and trigonometry » SL 3.2—2d and 3d trig
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Topic 3—Geometry and trigonometry

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