IBDP Maths AI: Topic SL 3.3: Applications of right and non-right angled trigonometry: IB style Questions HL Paper 1

Question

The diagram below shows a helicopter hovering at point H, 380 m vertically above a lake. Point A is the point on the surface of the lake, directly below the helicopter.

                                                                                            diagram not to scale

Minta is swimming at a constant speed in the direction of point A. Minta observes the helicopter from point C as she looks upward at an angle of 25°. After 15 minutes, Minta is at point B and she observes the same helicopter at an angle of 40°.

Find the distance from A to C. [2]

Find the distance from B to C. [3]

Find Minta’s speed, in metres per hour. [1]

▶️Answer/Explanation

(a) AC= \(\frac{380}{tan25°}\) OR   \(AC= \sqrt{\frac{380}{sin25°}-380^{2}}\)  OR

\(\frac{380}{tan25°}=\frac{AC}{sin65°}\)\)

\(AC = 815 M (814.912…)\)

(b)  AB = \(\frac{280}{tan40°}\)

= 453m (452.866..)

BC= 814.912…-452.866..

METHOD 2 attempt to find HB

\(HB =\frac{380}{sin40}\rightarrow
591m (=591.175…) \)

\(BC = \frac{591.175…\times sin15°}{sin25°}\)

\(= 362m (362.0.46..)\)

(c) \(362.046..\times 4 = 1450mh(1448.18..)\)

Question

Ellis designs a gift box. The top of the gift box is in the shape of a right-angled triangle GIK. A rectangular section HIJL is inscribed inside this triangle. The lengths of GH, JK, HL,

and LJ are p cm, q cm, 8 cm and 6 cm respectively.

                 

The area of the top of the gift box is A cm2.

a (i) Find A in terms of p and q .

(ii) Show that A = \(\frac{192}{q}\) + 3q + 48. [4]

(b) Find \(\frac{dA}{dq}\) [2]

Ellis wishes to find the value of q that will minimize the area of the top of the gift box.

(c) (i) Write down an equation Ellis could solve to find this value of q .

(ii) Hence, or otherwise, find this value of q . [2]

▶️Answer/Explanation

(a) (i) A = \(\frac{1}{2}\times6\times q + \frac{1}{2}\times8 \times p + 48\) OR A = \(\frac{1}{2}(p+6)(q+8)\) OR  A = 3q+4q + 48 (ii) valid attempt to link p and q, using tangents, similar triangles or other method eg. tan Θ = \(\frac{8}{p}\) and tan Θ = \(\frac{q}{6}\) OR tan Θ = \(\frac{p}{8}\) tan Θ = \(\frac{6}{q}\) OR \(\frac{8}{9}= \frac{q}{6}\) correct equation linking p and q eg. p q = 48 OR  p = \(\frac{48}{q}\) OR q = \(\frac{48}{p}\)  substitute p = \(\frac{48}{q}\) into a correct area expression eg. (A =) \(\frac{1}{2}\times6\times q + \frac{1}{2}\times8 \times p + 48\) OR (A =) \(\frac{1}{2}(\frac{48}{q}+ 6)(q+8)\) A = 3q + 48 (b) \(\frac{-192}{q^{2}}+3\) (c) (i) \(\frac{-192}{q^{2}}+3\) = 0 (ii) q = 8 cm

Question 6. [Maximum mark: 5]

A triangular field ABC is such that AB = 56 m and BC = 82 m , each measured correct to the nearest metre, and the angle at B is equal to 105, measured correct to the nearest 5′.

                                               

Calculate the maximum possible area of the field.

▶️Answer/Explanation

attempt to find any relevant maximum value largest sides are 56.5 and 82.5  smallest possible angle is 102.5  attempt to substitute into area of a triangle formula \(\frac{1}{2}\)×56.5× 82.5 × sin (102.5) = 2280m2 (2275.37…)

Question

The owner of a convenience store installs two security cameras, represented by points C1
and C2. Both cameras point towards the centre of the store’s cash register, represented by
the point R.
The following diagram shows this information on a cross-section of the store.

The cameras are positioned at a height of 3.1m, and the horizontal distance between the
cameras is 6.4m. The cash register is sitting on a counter so that its centre, R, is 1.0m
above the floor.
The distance from Camera 1 to the centre of the cash register is 2.8m.
(a) Determine the angle of depression from Camera 1 to the centre of the cash register.
Give your answer in degrees.
(b) Calculate the distance from Camera 2 to the centre of the cash register.
(c) Without further calculation, determine which camera has the largest angle of
depression to the centre of the cash register. Justify your response.

▶️Answer/Explanation

Ans:

(a) sin \(\theta = \frac{2.1}{2.8}\) OR tan \(\theta =\frac{2.1}{1.85202…}\)
\((\theta=) 48.6^0\)   \((48.5903…^0)\)
(b) METHOD 1
\(\sqrt{2.8^2-2.1^2}\)   OR   2.8cos (48.5903…)   OR \(\frac{2.1}{tan(48.5903…)}\)
1.85 m (1.85202…)
(6.4 – 1.85202…)
4.55 m (4.54797…)
\(\sqrt{(4.54797…)^2 + 2.1^2}\)
5.01 m (5.00939…m)

METHOD 2
attempt to use cosine rule
\((c^2=) 2.8^2 + 6.4^2 – 2(2.8)(6.4)cos(48.5903…)\)
(c=) 5.01 m   (5.00939…m)
(c) camera 1 is closer to the cash register than camera 2 (and both cameras are at the same height on the wall)
the larger angle of depression is from camera 1

Question

A new triangle DEF is positioned within a circle radius R such that DF is a diameter as shown in the following diagram.

a.i.In a triangle ABC, prove \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}\).[4]

a.ii.Prove that the area of the triangle ABC is \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\).[2]

a.iii.Given that R denotes the radius of the circumscribed circle prove that \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R\).[2]

a.iv.Hence show that the area of the triangle ABC is \(\frac{{abc}}{{4R}}\).[2]

b.i.Find in terms of R, the two values of (DE)2 such that the area of the shaded region is twice the area of the triangle DEF.[9]

b.ii.Using two diagrams, explain why there are two values of (DE)2.[2]

c.A parallelogram is positioned inside a circle such that all four vertices lie on the circle. Prove that it is a rectangle.[3]

▶️Answer/Explanation

Markscheme

\({\text{sin}}\,{\text{B}} = \frac{h}{c}\) and \({\text{sin}}\,{\text{C}} = \frac{h}{b}\) M1A1

hence \(h = c\,{\text{sin}}\,{\text{B}} = b\,{\text{sin}}\,{\text{C}}\) A1

by dropping a perpendicular from B, in exactly the same way we find \(c\,{\text{sin}}\,{\text{A}} = a\,{\text{sin}}\,{\text{C}}\) R1

hence \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}}\)

[4 marks]

a.i.

area = \(\frac{1}{2}ah\) M1A1

= \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\) AG

[2 marks]

a.ii.

since the angle at the centre of circle is twice the angle at the circumference \({\text{sin}}\,A = \frac{a}{{2R}}\) M1A1

hence \(\frac{a}{{{\text{sin}}\,A}} = 2R\) and therefore \(\frac{a}{{{\text{sin}}\,A}} = \frac{b}{{{\text{sin}}\,B}} = \frac{c}{{{\text{sin}}\,C}} = 2R\) AG

[2 marks]

a.iii.

area of the triangle is \(\frac{1}{2}ab\,{\text{sin}}\,{\text{C}}\) M1

since \({\text{sin}}\,{\text{C}} = \frac{c}{{2R}}\) A1

area of the triangle is \(\frac{1}{2}ab\,\frac{c}{{2R}} = \frac{{abc}}{{4R}}\) AG

[2 marks]

a.iv.

area of the triangle is \(\frac{{\pi {R^2}}}{6}\) (M1)A1

(DE)2 + (EF)2 = 4R2 M1

(DE)2 = 4R2 − (EF)2

\(\frac{1}{2}\left( {{\text{DE}}} \right)\left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{6} \Rightarrow \left( {{\text{EF}}} \right) = \frac{{\pi {R^2}}}{{3\left( {{\text{DE}}} \right)}}\) M1A1

\({\left( {{\text{DE}}} \right)^2} = 4{R^2} – \frac{{{\pi ^2}{R^4}}}{{9{{\left( {{\text{DE}}} \right)}^2}}}\) A1

\(9{\left( {{\text{DE}}} \right)^4} – 36{\left( {{\text{DE}}} \right)^2}{R^2} + {\pi ^2}{R^4} = 0\) A1

\({\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm \sqrt {1296{R^4} – 36{\pi ^2}{R^4}} }}{{18}}\) M1

\({\left( {{\text{DE}}} \right)^2} = \frac{{36{R^2} \pm 6{R^2}\sqrt {36 – {\pi ^2}} }}{{18}}\left( { = \frac{{6{R^2} \pm {R^2}\sqrt {36 – {\pi ^2}} }}{3}} \right)\) A1

[9 marks]

b.i.

 A1A1

[2 marks]

b.ii.

\(\mathop {\text{A}}\limits^ \wedge + \mathop {\text{C}}\limits^ \wedge = 180^\circ \) (cyclic quadrilateral) R1

however \(\mathop {\text{A}}\limits^ \wedge = \mathop {\text{C}}\limits^ \wedge \) (ABCD is a parallelogram) R1

\(\mathop {\text{A}}\limits^ \wedge = \mathop {\text{C}}\limits^ \wedge = 90^\circ \) A1

\(\mathop {\text{B}}\limits^ \wedge = \mathop {\text{D}}\limits^ \wedge = 90^\circ \)

hence ABCD is a rectangle AG

[3 marks]

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