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Geometry and Trigonometry — Free Maths AA HL Practice Questions

1FoundationSAQ-SEquation of a straight lines5 marksPaper 1~8 min
A straight line L1L_1 passes through the points A(2,5)A(2, 5) and B(6,13)B(6, 13).
(a)
Find the gradient of L1L_1[2 marks]
(b)
The perpendicular bisector of ABAB is the line L2L_2. Find the equation of L2L_2 in the form y=mx+cy = mx + c[3 marks]
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2MasterySAQ-SEquation of a straight lines8 marksPaper 1~12 min
A line LL has equation 2x+3y=122x + 3y = 12.
(a)
Find the gradient of LL and hence write down the gradient of any line perpendicular to LL[2 marks]
(b)
The line LL intersects the yy-axis at point BB and the xx-axis at point AA. Calculate the area of triangle OABOAB, where OO is the origin. [3 marks]
(c)
Find the perpendicular distance from OO to LL and verify that is consistent with the area found in part (b). [3 marks]
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3ChallengeSAQ-LEquation of a straight lines8 marksPaper 1~12 min
A line L1L_1 passes through point A(2,5)A(2,\,5) and has slope m1=3m_1 = -3.
(a)
Determine the equation of L1L_1 in the form ax+by+c=0ax + by + c = 0, where a,b,cZa, b, c \in \mathbb{Z}. Another line L2L_2 is perpendicular to L1L_1 and passes through point B(1,1)B(1,\,-1)[2 marks]
(b)
Determine the equation of L2L_2 in the form y=mx+cy = mx + c. The lines L1L_1 and L2L_2 intersect at point CC[2 marks]
(c)
Determine the coordinates of CC, giving your answer in exact form. [2 marks]
(d)
Determine the ratio AC:BCAC : BC, giving your answer in its simplest integer form. [2 marks]
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4FoundationSAQ-SEquation of a straight lines7 marksPaper 1~11 min
A line LL has equation 3x+4y=123x + 4y = 12.
(a)
Write LL in the form y=mx+cy = mx + c, stating the values of mm and cc[2 marks]
(b)
Find the coordinates of the xx-intercept of LL[1 mark]
(c)
A line MM is perpendicular to LL and passes through the point P(0,5)P(0,\,5). (i) Find the equation of MM in the form y=mx+cy = mx + c. [2]
(ii) Find the coordinates of the point of intersection of LL and MM[2 marks]
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5FoundationSAQ-SUnit circle and angle measurements5 marksPaper 1~8 min
A point PP lies on the unit circle such that the angle θ\theta between the positive xx-axis and the line OPOP is 5π6\dfrac{5\pi}{6} radians, measured anticlockwise.
(a)
State the coordinates of PP[2 marks]
(b)
Using your answer to part (a) and an appropriate compound-angle identity, find the exact value of cos(5π3)\cos\left(\dfrac{5\pi}{3}\right)[3 marks]
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6MasterySAQ-SDefinitions of sine, cosine, and tangents5 marksPaper 1~8 min
A particle moves such that its displacement, ss metres, from a fixed point OO at time tt seconds is given by s(t)=5cos ⁣(πt3)+12sin ⁣(πt3),t0.s(t) = 5\cos\!\left(\frac{\pi t}{3}\right) + 12\sin\!\left(\frac{\pi t}{3}\right), \quad t \geq 0.
(a)
Show that s(t)s(t) can be written in the form s(t)=Rcos ⁣(πt3α)s(t) = R\cos\!\left(\dfrac{\pi t}{3} - \alpha\right), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}. State the value of RR and the exact value of α\alpha in the form arctan(k)\arctan(k)[3 marks]
(b)
Hence find the first time t>0t > 0 at which the particle is at OO, giving your answer in the form a+bπarctan(k)a + \dfrac{b}{\pi}\arctan(k), where a,b,ka, b, k are constants to be determined. [2 marks]
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7ChallengeSAQ-LDefinitions of sine, cosine, and tangents8 marksPaper 1~12 min
A surveyor measures the angle of elevation to the top of a vertical cliff from two points on level ground. From point AA, the angle of elevation is π6\dfrac{\pi}{6} radians. From point BB, which is 120120 metres closer to the base of the cliff, the angle of elevation is π3\dfrac{\pi}{3} radians. Let dd be the horizontal distance from BB to the base of the cliff, and let hh be the height of the cliff.
(a)
Show that h=d3h = d\sqrt{3} and hence determine the exact value of hh[4 marks]
(b)
The surveyor now considers a general scenario in which the angle of elevation from AA is α\alpha radians, where 0<α<π30 < \alpha < \dfrac{\pi}{3}, the angle from BB remains π3\dfrac{\pi}{3}, and the distance ABAB remains 120120 m. Show that the height of the cliff is given by h=1203tanα3tanαh = \frac{120\sqrt{3}\tan\alpha}{\sqrt{3} - \tan\alpha} and hence deduce, with justification, whether hh is an increasing or decreasing function of α\alpha on the given domain. [4 marks]
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8FoundationSAQ-SUnit circle and angle measurements5 marksPaper 1~8 min
A point QQ on the unit circle has coordinates (22, 22)\left(\dfrac{\sqrt{2}}{2},\ -\dfrac{\sqrt{2}}{2}\right), where θ\theta is the angle measured anticlockwise from the positive xx-axis to OQOQ, and 0θ<2π0 \leq \theta < 2\pi.
(a)
State the exact value of sinθ\sin\theta[1 mark]
(b)
Find the value of θ\theta in radians. [2 marks]
(c)
Find the exact value of sin ⁣(θ+3π4)\sin\!\left(\theta + \dfrac{3\pi}{4}\right)[2 marks]
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9FoundationSAQ-SLaw of Sines and its applicationss5 marksPaper 1~8 min
A surveyor needs to find the height of a vertical cliff. From a point PP on the ground, the angle of elevation to the top of the cliff is 3030^\circ. The surveyor then walks 100m100\,\text{m} directly towards the cliff to a point QQ. From QQ, the angle of elevation to the top of the cliff is 4545^\circ. Let the height of the cliff be hh metres and let the horizontal distance from PP to the base of the cliff be dPd_P metres.
(a)
Show that dP=h3d_P = h\sqrt{3}[2 marks]
(b)
Write down the horizontal distance from QQ to the base of the cliff in terms of hh[1 mark]
(c)
Hence, find the value of hh, giving your answer in the form a+b3a + b\sqrt{3} where a,bZa, b \in \mathbb{Z}[2 marks]
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10MasterySAQ-SLaw of Sines and its applicationss5 marksPaper 1~8 min
In triangle PQRPQR, PQ=10cmPQ = 10\,\text{cm}, PQR=42\angle PQR = 42^\circ, and PRQ=68\angle PRQ = 68^\circ.
(a)
Show that PR=10sin42sin68PR = \dfrac{10\sin 42^\circ}{\sin 68^\circ}[2 marks]
(b)
Using the identity sin68=cos22\sin 68^\circ = \cos 22^\circ and the result from part (a), show that PR=10sin42cos22PR = \dfrac{10\sin 42^\circ}{\cos 22^\circ}[1 mark]
(c)
Hence, using an appropriate double-angle identity, show that PR=5sin42sin11cos11PR = \dfrac{5\sin 42^\circ}{\sin 11^\circ \cos 11^\circ}, and deduce an expression for PRPR in the form ksin42sin22\dfrac{k\sin 42^\circ}{\sin 22^\circ}, stating the value of kk[2 marks]
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11ChallengeSAQ-LLaw of Sines and its applicationss8 marksPaper 1~12 min
In triangle ABCABC, side AB=10cmAB = 10\,\text{cm}, side BC=8cmBC = 8\,\text{cm}, and BAC=θ\angle BAC = \theta, where θ\theta is acute and sinθ=45\sin\theta = \dfrac{4}{5}.
(a)
Determine the two possible values of ACB\angle ACB, giving your answers in degrees correct to one decimal place. [4 marks]
(b)
For each value of ACB\angle ACB found in part (a), determine the corresponding length of ACAC, giving your answers in cm correct to three significant figures. [2 marks]
(c)
Analyse whether the two possible triangles have the same area. Support your answer with appropriate calculations. [2 marks]
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12FoundationSAQ-SLaw of Sines and its applicationss7 marksPaper 1~11 min
In triangle ABCABC, side a=8cma = 8\,\text{cm}, side b=6cmb = 6\,\text{cm}, and angle A=60°A = 60°.
(a)
Use the Law of Sines to show that sinB=338\sin B = \dfrac{3\sqrt{3}}{8}[2 marks]
(b)
Find the two possible values of angle BB, correct to 11 decimal place. [2 marks]
(c)
For the case in which triangle ABCABC is acute, find the exact length of side cc[3 marks]
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13MasterySAQ-SEquation of a straight lines6 marksPaper 2~9 min
A straight line L1L_1 passes through the points A(2,5)A(2,\,5) and B(4,11)B(4,\,11).
(a)
Show that the equation of L1L_1 is y=3x1y = 3x - 1[2 marks]
(b)
A second line L2L_2 is perpendicular to L1L_1 and passes through the midpoint MM of ABAB. Determine the equation of L2L_2 in the form y=mx+cy = mx + c[2 marks]
(c)
Find the coordinates of the point of intersection of L1L_1 and L2L_2, and hence verify that this point is MM[2 marks]
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14MasterySAQ-SEquation of a straight lines6 marksPaper 2~9 min
An aircraft flies in a straight line at constant speed. At time t=0st = 0\,\text{s}, its position is P(100,200)mP(100,\,200)\,\text{m} relative to a control tower at the origin. At t=10st = 10\,\text{s}, its position is Q(250,500)mQ(250,\,500)\,\text{m}.
(a)
Calculate the slope of the flight path PQPQ[2 marks]
(b)
Determine the equation of line PQPQ in the form ax+by=cax + by = c, where a,b,cZa,b,c \in \mathbb{Z}[2 marks]
(c)
The aircraft's position at time tt seconds is given by r(t)=(100200)+t(1530).\mathbf{r}(t) = \begin{pmatrix} 100 \\ 200 \end{pmatrix} + t\begin{pmatrix} 15 \\ 30 \end{pmatrix}. Two radio beacons are located at A(700,1400)mA(700,\,1400)\,\text{m} and B(1150,1700)mB(1150,\,1700)\,\text{m}. Determine the time tt at which the aircraft is equidistant from AA and BB[2 marks]
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15MasterySAQ-SEquation of a straight lines6 marksPaper 2~9 min
A ski lift a resort follows a straight-line path from the base station at point A(20,50)A(20, 50) to the summit at point B(120,200)B(120, 200), where coordinates are in metres.
(a)
Calculate the gradient of the line ABAB, giving your answer as a simplified fraction. [2 marks]
(b)
A maintenance hut is located at point C(60,k)C(60, k) on the line ABAB. Determine the value of kk[2 marks]
(c)
A second resort claims its ski lift, running along the line y=23x+1903y = -\dfrac{2}{3}x + \dfrac{190}{3}, passes through the base station A(20,50)A(20, 50) and is perpendicular to lift ABAB. A skier starts at AA and travels along this second lift until reaching the point DD directly above CC (i.e. DD has the same xx-coordinate as CC). Determine the length ADAD and hence evaluate whether DD lies closer to AA than CC does. [2 marks]
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16MasterySAQ-SDistance formula, midpoint formula, and area of triangles6 marksPaper 2~9 min
A farmer has a triangular field ABCABC. The vertices are located at A(2,5)A(2,5), B(8,1)B(8,1), and C(6,9)C(6,9), where distances are measured in kilometres.
(a)
Calculate the length of side ABAB, giving your answer in exact form. [2 marks]
(b)
Find the coordinates of the midpoint MM of side ACAC[1 mark]
(c)
Calculate the area of triangle BCMBCM with vertices B(8,1)B(8,1), C(6,9)C(6,9), M(4,7)M(4,7), giving your answer in km2\text{km}^2[2 marks]
(d)
Hence, deduce the area of triangle ABMABM and explain what this result reveals about the median BMBM of triangle ABCABC. Formulae provided: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} Midpoint=(x1+x22,y1+y22)\text{Midpoint} = \left(\frac{x_1+x_2}{2},\,\frac{y_1+y_2}{2}\right) Area=12x1(y2y3)+x2(y3y1)+x3(y1y2)\text{Area} = \tfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right| [1 mark]
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17MasterySAQ-SDefinitions of sine, cosine, and tangents6 marksPaper 2~9 min
A vertical radio mast of height 30m30\,\text{m} is supported by two straight cables attached to the top of the mast and anchored at points AA and BB on horizontal ground on opposite sides of the mast. The cable to AA makes angle of 25°25° with the horizontal; the cable to BB makes angle of 40°40° with the horizontal.
(a)
Calculate the distance from the base of the mast to anchor point AA. Give your answer correct to 3 significant figures. [2 marks]
(b)
Calculate the length of the cable to anchor point BB. Give your answer correct to 3 significant figures. [2 marks]
(c)
Given that AA and BB lie on opposite sides of the mast, calculate the distance ABAB. Give your answer correct to 3 significant figures. [2 marks]
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18MasterySAQ-SDefinitions of sine, cosine, and tangents6 marksPaper 2~9 min
A surveyor measures the angle of elevation to the top of a tower TT from two points AA and BB on horizontal ground. Points AA, BB, and the base of the tower are collinear. From point AA, the angle of elevation is 32°32°. From point BB, which is 50m50\,\text{m} closer to the tower than AA, the angle of elevation is 48°48°.
(a)
Let the height of the tower be hh metres and the horizontal distance from BB to the base of the tower be xx metres. Show that h=xtan(48°)andh=(x+50)tan(32°).h = x\tan(48°) \quad \text{and} \quad h = (x+50)\tan(32°). [2 marks]
(b)
Hence, determine the value of xx. Give your answer correct to 3 significant figures. [2 marks]
(c)
Calculate the straight-line distance from point AA to the top of the tower TT. Give your answer correct to 3 significant figures. [2 marks]
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19MasterySAQ-SDefinitions of sine, cosine, and tangents6 marksPaper 2~9 min
A Ferris wheel at a theme park has a diameter of 4040 metres. The bottom of the wheel is 22 metres above the ground. The wheel completes one full revolution every 6060 seconds. A rider boards the wheel at the lowest point at time t=0t = 0 seconds.
(a)
Show that the height hh metres of the rider above the ground at time tt seconds can be modelled by h(t)=2220cos ⁣(πt30).h(t) = 22 - 20\cos\!\left(\frac{\pi t}{30}\right). [2 marks]
(b)
Calculate the first time at which the rider is 3434 metres above the ground. Give your answer correct to 3 significant figures. [4 marks]
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20MasterySAQ-SDefinitions of sine, cosine, and tangents6 marksPaper 2~9 min
A surveyor measures the angle of elevation to the top of a vertical tower from point A on level ground as 35°35°. She then walks 5050 metres directly towards the tower to point B, where the angle of elevation is 52°52°. Let the height of the tower be hh metres and the distance from point B to the base of the tower be xx metres.
(a)
Show that h=xtan52°h = x\tan 52° and h=(x+50)tan35°h = (x + 50)\tan 35°[2 marks]
(b)
Hence show that x=50tan35°tan52°tan35°x = \dfrac{50\tan 35°}{\tan 52° - \tan 35°}[2 marks]
(c)
Calculate the height of the tower, giving your answer in metres correct to 3 significant figures. [2 marks]
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21MasterySAQ-SLaw of Sines and its applicationss6 marksPaper 2~9 min
A surveyor determines the height of a vertical cliff face TCTC, where CC is the base of the cliff at ground level. From point AA on level ground, the angle of elevation to the top of the cliff TT is 27.527.5^\circ. The surveyor walks 120m120\,\text{m} directly towards the cliff to point BB. From BB, the angle of elevation to TT is 41.241.2^\circ.
(a)
Show that the angle ATB=13.7\angle ATB = 13.7^\circ[2 marks]
(b)
Calculate the height hh of the cliff, giving your answer correct to 3 significant figures. [4 marks]
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22MasterySAQ-SLaw of Sines and its applicationss6 marksPaper 2~9 min
A ship sails from port OO to point AA, a distance of 25km25\,\text{km} on a bearing of 030°030°. From AA, the ship sails to point BB on a bearing of 110°110°. The distance OB=40kmOB = 40\,\text{km}.
(a)
Show that angle OAB=80°OAB = 80°[2 marks]
(b)
Calculate the distance ABAB, giving your answer correct to 3 significant figures. [2 marks]
(c)
A lighthouse LL is located such that is equidistant from OO, AA, and BB. Determine the bearing of LL from OO, correct to the nearest degree. [2 marks]
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23MasterySAQ-SLaw of Sines and its applicationss6 marksPaper 2~9 min
A triangular garden ABCABC has AB=15mAB = 15\,\text{m}, AC=20mAC = 20\,\text{m}, and BAC=72°\angle BAC = 72°. A straight path runs from vertex BB to the midpoint MM of side ACAC.
(a)
Calculate the length BCBC, giving your answer correct to 3 significant figures. [2 marks]
(b)
Hence, use the law of sines to find ACB\angle ACB, giving your answer correct to 1 decimal place. [2 marks]
(c)
Determine the length of the path BMBM, giving your answer correct to 3 significant figures. [2 marks]
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24MasterySAQ-SLaw of Sines and its applicationss4 marksPaper 2~6 min
A surveyor determines the distance across a river from point AA one bank to point BB on the opposite bank. Point CC is chosen on the same bank as AA such that AC=85.0mAC = 85.0\,\text{m}. The surveyor measures BAC=62.4°\angle BAC = 62.4° and ACB=48.7°\angle ACB = 48.7°.
(a)
Calculate the measure of ABC\angle ABC[1 mark]
(b)
Hence, determine the distance ABAB. Give your answer correct to 3 significant figures. [3 marks]
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25ChallengeLAQEquation of a straight lines10 marksPaper 3~15 min
A line L1L_1 passes through the points A(1,3)A(1,3) and B(5,11)B(5,11). A second line L2L_2 is perpendicular to L1L_1 and passes through the point C(2,1)C(2,-1). A third line L3L_3 is the reflection of L2L_2 in the line y=xy = x.
(a)
Show that the equation of L1L_1 is y=2x+1y = 2x + 1[3 marks]
(b)
(i) Find the equation of L2L_2 in the form ax+by+c=0ax + by + c = 0, where a,b,cZa, b, c \in \mathbb{Z}. [3 marks]
(ii) Find the coordinates of the point DD where L1L_1 and L2L_2 intersect. [2 marks]
(c)
L3L_3 intersects L1L_1 at point FF. Determine the coordinates of FF and hence show that DD, CC, and FF are not collinear. [2 marks]
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26ChallengeLAQDistance formula, midpoint formula, and area of triangles10 marksPaper 3~15 min
Consider the points A(1,2)A(1,2), B(5,5)B(5,5), and C(3,7)C(3,7).
(a)
(i) Calculate the midpoint MM of ABAB and the midpoint NN of BCBC. [2 marks]
(ii) Show that MNACMN \parallel AC. [2 marks]
(iii) Prove that MN=12ACMN = \dfrac{1}{2}AC[2 marks]
(b)
A triangle has vertices P(p1,p2)P(p_1,p_2), Q(q1,q2)Q(q_1,q_2), and R(r1,r2)R(r_1,r_2). The midpoints of PQPQ, QRQR, and RPRP are LL, MM, and NN respectively. Prove that the area of triangle LMNLMN is one quarter of the area of triangle PQRPQR. The following formulae may be used: - *Midpoint of (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2): (x1+x22,y1+y22)\left(\dfrac{x_1+x_2}{2},\,\dfrac{y_1+y_2}{2}\right)* - *Distance: d=(x2x1)2+(y2y1)2d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}* - *Area of triangle with vertices (x1,y1)(x_1,y_1), (x2,y2)(x_2,y_2), (x3,y3)(x_3,y_3): 12x1(y2y3)+x2(y3y1)+x3(y1y2)\dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|[4 marks]
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27ChallengeLAQDefinitions of sine, cosine, and tangents14 marksPaper 3~21 min

Data

(ii)Hence,usingtheSqueezeTheorem,provethat(ii) Hence, using the Squeeze Theorem, prove that\lim_{x \to 0^+} \frac{\sin x}{x} = 1. \quad [4]$$ (iii) State why the result in (ii) extends to $\lim_{x \to 0} \dfrac{\sin x}{x} = 1$. $\quad [1] You may use without proof that cosx1\cos x \to 1 as x0x \to 0, and that sin2x+cos2x=1\sin^2 x + \cos^2 x = 1.
A function gg is defined by g(x)=sinxxg(x) = \dfrac{\sin x}{x} for x0x \neq 0.
(a)
Using the unit circle below, where 0<x<π20 < x < \dfrac{\pi}{2}, the following areas can be established: Area of OAPArea of sector OAPArea of OAT\text{Area of } \triangle OAP \leq \text{Area of sector } OAP \leq \text{Area of } \triangle OAT where OO is the origin, A=(1,0)A = (1,0), P=(cosx,sinx)P = (\cos x, \sin x) lies on the unit circle, and TT is the point where the tangent to the circle at AA meets the line OPOP extended. (i) Show that the three areas satisfy $$\frac{1}{2}\sin x \leq \frac{x}{2} \leq \frac{1}{2}\tan x. \quad [3 marks]
(b)
Using the result from (a), evaluate limx01cosxx2,\lim_{x \to 0} \frac{1 - \cos x}{x^2}, justifying each step of your reasoning. $\quad [6 marks]
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28ChallengeLAQUnit circle and angle measurements15 marksPaper 3~23 min
Consider a point PP on the unit circle, defined by the angle θ\theta measured anticlockwise from the positive xx-axis, with coordinates (cosθ,sinθ)(\cos\theta, \sin\theta).
(a)
Prove that for any angle θ\theta: cos ⁣(θ+π3)cos ⁣(θπ3)=cos2θ34.\cos\!\left(\theta+\frac{\pi}{3}\right)\cos\!\left(\theta-\frac{\pi}{3}\right)=\cos^2\theta-\frac{3}{4}. [5 marks]
(b)
A particle PP moves on the unit circle with position at time tt seconds given by P(t)=(cos ⁣(π3t),sin ⁣(π3t)).P(t)=\left(\cos\!\left(\frac{\pi}{3}t\right), \sin\!\left(\frac{\pi}{3}t\right)\right). A second particle QQ moves on the same circle with position Q(t)=(cos ⁣(π3t+π6),sin ⁣(π3t+π6)).Q(t)=\left(\cos\!\left(\frac{\pi}{3}t+\frac{\pi}{6}\right), \sin\!\left(\frac{\pi}{3}t+\frac{\pi}{6}\right)\right). (i) Determine the value of tt in the interval 0t30 \leq t \leq 3 for which the xx-coordinates of PP and QQ are equal. Justify your reasoning. [3 marks]
(ii) Show that the distance between P(t)P(t) and Q(t)Q(t) is constant for all tt, and state its exact value. [2 marks]
(iii) The xx-coordinate of PP at time tt is xP=cos ⁣(π3t)x_P = \cos\!\left(\dfrac{\pi}{3}t\right). Using the result from part (a), find all values of tt in 0t60 \leq t \leq 6 for which xP(t+1)xP(t1)=12,x_P(t+1) \cdot x_P(t-1) = -\frac{1}{2}, and interpret geometrically what this condition means for the position of PP on the unit circle. [5 marks]
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29ChallengeLAQLaw of Sines and its applicationss10 marksPaper 3~15 min
In a triangle ABCABC, let aa, bb, cc denote the lengths of sides opposite vertices AA, BB, CC respectively, and let RR denote the circumradius. A surveyor measures a triangular plot of land PQRPQR. The angle QPR=75\angle QPR = 75^\circ, the side PQ=120mPQ = 120\,\text{m}, and the angle PQR=60\angle PQR = 60^\circ.
(a)
Prove that for any triangle ABCABC: asinA=bsinB=csinC=2R\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R where RR is the circumradius. [5 marks]
(b)
Using the result from part (a), find the exact values of PRPR and the circumradius RR of triangle PQRPQR[3 marks]
(c)
A colleague suggests that the surveyor could instead have measured PQR=45\angle PQR = 45^\circ, with PQ=120mPQ = 120\,\text{m} and QR=606mQR = 60\sqrt{6}\,\text{m} unchanged. Determine the number of distinct triangles PQRPQR satisfying these three measurements, and justify your answer. [2 marks]
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30ChallengeLAQLaw of Sines and its applicationss10 marksPaper 3~15 min
A lighthouse LL is located off a straight coastline. Two observation points AA and BB are on the coastline, with AB=500mAB = 500\,\text{m}. From AA, the bearing of LL is 040°040°. From BB, the bearing of LL is 320°320°. Let dd be the perpendicular distance from LL to the line ABAB.
(a)
Show that the interior angles of triangle LABLAB at vertices AA and BB are each 50°50°, and hence find the angle ALB\angle ALB[3 marks]
(b)
Using the Law of Sines, find ALAL in exact form. Hence show that d=500sin250°sin80°m.d = \frac{500\sin^2 50°}{\sin 80°}\,\text{m.} [4 marks]
(c)
A third observation point CC is placed on the coastline between AA and BB such that AC=200mAC = 200\,\text{m}. The bearing of LL from CC is θ\theta, where 0°<θ<90°0° < \theta < 90°. (i) Show that tanθ=dCLx\tan\theta = \dfrac{d}{CL_x}, where CLxCL_x is the horizontal distance from CC to the foot of the perpendicular from LL to ABAB. Express CLxCL_x in terms of known lengths. [1]
(ii) Hence find the exact value of tanθ\tan\theta, and determine whether θ>40°\theta > 40° or θ<40°\theta < 40°. Justify your answer. [2 marks]
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