Home / IB DP Math AA Topic SL 3.5 :Definition of cosθ , sinθ and tanθ in terms of the unit circle HL Paper 2

IB DP Math AA Topic SL 3.5 :Definition of cosθ , sinθ and tanθ in terms of the unit circle HL Paper 2

Question

Consider a triangle ABC with \({\rm{B\hat AC}} = 45.7^\circ \) , AB = 9.63 cm and BC = 7.5 cm .

a.By drawing a diagram, show why there are two triangles consistent with this information.[2]

 

b.Find the possible values of AC .[6]

 
▶️Answer/Explanation

Markscheme

     A2

Note: Accept 2 separate triangles. The diagram(s) should show that one triangle has an acute angle and the other triangle has an obtuse angle. The values 9.63, 7.5 and 45.7 and/or the letters, A, B, C′ and C should be correctly marked on the diagram(s).

 

[2 marks]

a.

METHOD 1

\(\frac{{\sin 45.7}}{{7.5}} = \frac{{\sin C}}{{9.63}}\)     M1

\( \Rightarrow \hat C = 66.77…^\circ \,{\text{, }}113.2…^\circ \)     (A1)(A1)

\( \Rightarrow \hat B = 67.52…^\circ \,{\text{, }}21.07…^\circ \)     (A1)

\(\frac{b}{{\sin B}} = \frac{{7.5}}{{\sin 45.7}} \Rightarrow b = 9.68({\text{cm}}){\text{, }}b = 3.77({\text{cm}})\)     A1A1 

Note: If only the acute value of \({\hat C}\) is found, award M1(A1)(A0)(A0)A1A0.

 

METHOD 2

\({7.5^2} = {9.63^2} + {b^2} – 2 \times 9.63 \times b\cos 45.7^\circ \)     M1A1

\({b^2} – 13.45…b + 36.48… = 0\)

\(b = \frac{{13.45… \pm \sqrt {13.45..{.^2} – 4 \times 36.48…} }}{2}\)     (M1)(A1)

\({\text{AC}} = 9.68({\text{cm}})\,{\text{, AC}} = 3.77({\text{cm}})\)     A1A1

[6 marks]

b.

Question

The diagram shows the plan of an art gallery a metres wide. [AB] represents a doorway, leading to an exit corridor b metres wide. In order to remove a painting from the art gallery, CD (denoted by L ) is measured for various values of \(\alpha \) , as represented in the diagram.

a.If \(\alpha \) is the angle between [CD] and the wall, show that \(L = \frac{a }{{\sin \alpha }} + \frac{b}{{\cos \alpha }}{\text{, }}0 < \alpha  < \frac{\pi }{2}\).[3]

 

b.If a = 5 and b = 1, find the maximum length of a painting that can be removed through this doorway.[4]

 

Let a = 3k and b = k .

c.Find \(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }}\).[3]

 

Let a = 3k and b = k

d.Find, in terms of k , the maximum length of a painting that can be removed from the gallery through this doorway.[6]

 

e.Let a = 3k and b = k

Find the minimum value of k if a painting 8 metres long is to be removed through this doorway.

[2]
 
▶️Answer/Explanation

Markscheme

\(L = {\text{CA}} + {\text{AD}}\)     M1

\({\text{sin}}\alpha {\text{ = }}\frac{a}{{{\text{CA}}}} \Rightarrow {\text{CA}} = \frac{a}{{\sin \alpha }}\)     A1

\(\cos \alpha  = \frac{b}{{{\text{AD}}}} \Rightarrow {\text{AD}} = \frac{b}{{\cos \alpha }}\)     A1

\(L = \frac{a}{{\sin \alpha }} + \frac{b}{{\cos \alpha }}\)     AG

[2 marks]

a.

\(a = 5{\text{ and }}b = 1 \Rightarrow L = \frac{5}{{\sin \alpha }} + \frac{1}{{\cos \alpha }}\)

 

METHOD 1

     (M1)

minimum from graph \( \Rightarrow L = 7.77\)     (M1)A1

minimum of gives the max length of the painting     R1

[4 marks]

METHOD 2

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 5\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{\sin \alpha }}{{{{\cos }^2}\alpha }}\)     (M1)

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = 5 \Rightarrow \tan \alpha  = \sqrt[{3{\text{ }}}]{5}{\text{ }}(\alpha  = 1.0416…)\)     (M1)

minimum of gives the max length of the painting     R1

maximum length = 7.77     A1

[4 marks]

b.

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k\cos \alpha }}{{{{\sin }^2}\alpha }} + \frac{{k\sin \alpha }}{{{{\cos }^2}\alpha }}\,\,\,\,\,{\text{(or equivalent)}}\)     M1A1A1

[3 marks]

c.

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = \frac{{ – 3k{{\cos }^3}\alpha  + k{{\sin }^3}\alpha }}{{{{\sin }^2}\alpha {{\cos }^2}\alpha }}\)     (A1)

\(\frac{{{\text{d}}L}}{{{\text{d}}\alpha }} = 0 \Rightarrow \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = \frac{{3k}}{k} \Rightarrow \tan \alpha  = \sqrt[3]{3}\,\,\,\,\,(\alpha  = 0.96454…)\)     M1A1

\(\tan \alpha  = \sqrt[3]{3} \Rightarrow \frac{1}{{\cos \alpha }} = \sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(1.755…)\)     (A1)

\({\text{and }}\frac{1}{{\sin \alpha }} = \frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}\,\,\,\,\,(1.216…)\)     (A1)

\(L = 3k\left( {\frac{{\sqrt {1 + \sqrt[3]{9}} }}{{\sqrt[3]{3}}}} \right) + k\sqrt {1 + \sqrt[3]{9}} \,\,\,\,\,(L = 5.405598…k)\)     A1     N4

[6 marks]

d.

\(L \leqslant 8 \Rightarrow k \geqslant 1.48\)     M1A1

the minimum value is 1.48

[2 marks]

e.

Examiners report

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

a.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

b.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

c.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

d.

Part (a) was very well done by most candidates. Parts (b), (c) and (d) required a subtle balance between abstraction, differentiation skills and use of GDC. 

In part (b), although candidates were asked to justify their reasoning, very few candidates offered an explanation for the maximum. Therefore most candidates did not earn the R1 mark in part (b). Also not as many candidates as anticipated used a graphical approach, preferring to use the calculus with varying degrees of success. In part (c), some candidates calculated the derivatives of inverse trigonometric functions. Some candidates had difficulty with parts (d) and (e). In part (d), some candidates erroneously used their alpha value from part (b). In part (d) many candidates used GDC to calculate decimal values for \(\alpha \) and L. The premature rounding of decimals led sometimes to inaccurate results. Nevertheless many candidates got excellent results in this question.

e.

Question

A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle \(\theta \) where \(\theta  = {\rm{A\hat PB}}\), as shown in the diagram.

a.Find an expression for \(\theta \) in terms of x, where x is the distance of P from the base of the wall of height 8 m.[2]

 

b.(i)     Calculate the value of \(\theta \) when x = 0.

(ii)     Calculate the value of \(\theta \) when x = 20.[2]

c.Sketch the graph of \(\theta \), for \(0 \leqslant x \leqslant 20\).[2]

d.Show that \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\).[6]

e.Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures.[3]

f.The point P moves across the street with speed \(0.5{\text{ m}}{{\text{s}}^{ – 1}}\). Determine the rate of change of \(\theta \) with respect to time when P is at the midpoint of the street [4]

 
▶️Answer/Explanation

Markscheme

EITHER

\(\theta  = \pi  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent)     M1A1

Note: Accept \(\theta  = 180^\circ  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent).

 

OR

\(\theta  = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\) (or equivalent)     M1A1

[2 marks]

a.

(i)     \(\theta  = 0.994{\text{ }}\left( { = \arctan \frac{{20}}{{13}}} \right)\)     A1

 

(ii)     \(\theta  = 1.19{\text{ }}\left( { = \arctan \frac{5}{2}} \right)\)     A1

[2 marks]

b.

correct shape.     A1

correct domain indicated.     A1

 

 

[2 marks]

c.

attempting to differentiate one \(\arctan \left( {f(x)} \right)\) term     M1

EITHER

\(\theta  = \pi  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\)

\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{8}{{{x^2}}} \times \frac{1}{{1 + {{\left( {\frac{8}{x}} \right)}^2}}} – \frac{{13}}{{{{(20 – x)}^2}}} \times \frac{1}{{1 + {{\left( {\frac{{13}}{{20 – x}}} \right)}^2}}}\)     A1A1

OR

\(\theta  = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\)

\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{\frac{1}{8}}}{{1 + {{\left( {\frac{x}{8}} \right)}^2}}} + \frac{{ – \frac{1}{{13}}}}{{1 + {{\left( {\frac{{20 – x}}{{13}}} \right)}^2}}}\)     A1A1

THEN

\( = \frac{8}{{{x^2} + 64}} – \frac{{13}}{{569 – 40x + {x^2}}}\)     A1

\( = \frac{{8(569 – 40x + {x^2}) – 13({x^{2}} + 64)}}{{({x^2} + 64)({x^2} – 40x + 569)}}\)     M1A1

\( = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\)     AG

[6 marks]

d.

Maximum light intensity at P occurs when \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\).     (M1)

either attempting to solve \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\) for x or using the graph of either \(\theta \) or \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}}\)     (M1)

x = 10.05 (m)     A1

[3 marks]

e.

\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5\)     (A1)

At x = 10, \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0.000453{\text{ }}\left( { = \frac{5}{{11029}}} \right)\).     (A1)

use of \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}x}}{{{\text{d}}t}}\)     M1

\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.000227{\text{ }}\left( { = \frac{5}{{22058}}} \right){\text{ (rad }}{{\text{s}}^{ – 1}})\)     A1

Note: Award (A1) for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} =  – 0.5\) and A1 for \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = – 0.000227{\text{ }}\left( { = – \frac{5}{{22058}}} \right){\text{ }}\).

Note: Implicit differentiation can be used to find \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\). Award as above.

[4 marks]

 

Question

A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle \(\theta \) where \(\theta  = {\rm{A\hat PB}}\), as shown in the diagram.

a.Find an expression for \(\theta \) in terms of x, where x is the distance of P from the base of the wall of height 8 m.[2]

 

b.(i)     Calculate the value of \(\theta \) when x = 0.

(ii)     Calculate the value of \(\theta \) when x = 20.[2]

 

c.Sketch the graph of \(\theta \), for \(0 \leqslant x \leqslant 20\).[2]

 

d.Show that \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\).[6]

 

e.Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures.[3]

 

f.The point P moves across the street with speed \(0.5{\text{ m}}{{\text{s}}^{ – 1}}\). Determine the rate of change of \(\theta \) with respect to time when P is at the midpoint of the street[4]

 
▶️Answer/Explanation

Markscheme

EITHER

\(\theta  = \pi  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent)     M1A1

Note: Accept \(\theta  = 180^\circ  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\) (or equivalent).

 

OR

\(\theta  = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\) (or equivalent)     M1A1

[2 marks]

a.

(i)     \(\theta  = 0.994{\text{ }}\left( { = \arctan \frac{{20}}{{13}}} \right)\)     A1

 

(ii)     \(\theta  = 1.19{\text{ }}\left( { = \arctan \frac{5}{2}} \right)\)     A1

[2 marks]

b.

correct shape.     A1

correct domain indicated.     A1

 

 

[2 marks]

c.

attempting to differentiate one \(\arctan \left( {f(x)} \right)\) term     M1

EITHER

\(\theta  = \pi  – \arctan \left( {\frac{8}{x}} \right) – \arctan \left( {\frac{{13}}{{20 – x}}} \right)\)

\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{8}{{{x^2}}} \times \frac{1}{{1 + {{\left( {\frac{8}{x}} \right)}^2}}} – \frac{{13}}{{{{(20 – x)}^2}}} \times \frac{1}{{1 + {{\left( {\frac{{13}}{{20 – x}}} \right)}^2}}}\)     A1A1

OR

\(\theta  = \arctan \left( {\frac{x}{8}} \right) + \arctan \left( {\frac{{20 – x}}{{13}}} \right)\)

\(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = \frac{{\frac{1}{8}}}{{1 + {{\left( {\frac{x}{8}} \right)}^2}}} + \frac{{ – \frac{1}{{13}}}}{{1 + {{\left( {\frac{{20 – x}}{{13}}} \right)}^2}}}\)     A1A1

THEN

\( = \frac{8}{{{x^2} + 64}} – \frac{{13}}{{569 – 40x + {x^2}}}\)     A1

\( = \frac{{8(569 – 40x + {x^2}) – 13({x^{2}} + 64)}}{{({x^2} + 64)({x^2} – 40x + 569)}}\)     M1A1

\( = \frac{{5(744 – 64x – {x^2})}}{{({x^2} + 64)({x^2} – 40x + 569)}}\)     AG

[6 marks]

d.

Maximum light intensity at P occurs when \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\).     (M1)

either attempting to solve \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0\) for x or using the graph of either \(\theta \) or \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}}\)     (M1)

x = 10.05 (m)     A1

[3 marks]

e.

\(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0.5\)     (A1)

At x = 10, \(\frac{{{\text{d}}\theta }}{{{\text{d}}x}} = 0.000453{\text{ }}\left( { = \frac{5}{{11029}}} \right)\).     (A1)

use of \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}x}} \times \frac{{{\text{d}}x}}{{{\text{d}}t}}\)     M1

\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.000227{\text{ }}\left( { = \frac{5}{{22058}}} \right){\text{ (rad }}{{\text{s}}^{ – 1}})\)     A1

Note: Award (A1) for \(\frac{{{\text{d}}x}}{{{\text{d}}t}} =  – 0.5\) and A1 for \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = – 0.000227{\text{ }}\left( { = – \frac{5}{{22058}}} \right){\text{ }}\).

Note: Implicit differentiation can be used to find \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\). Award as above.

[4 marks]

f.
Scroll to Top