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Geometry and Trigonometry — Free Maths AI HL Practice Questions

1FoundationSAQ-SEquation of a straight line and slope-intercept form5 marksPaper 1~8 min
The cost, CC dollars, of renting a bicycle is modelled by C=mt+cC = mt + c, where tt is the rental time in hours, mm is the hourly rate in dollars per hour, and cc is a fixed booking fee in dollars. For a 3-hour rental the cost is 22 USD. For a 7-hour rental the cost is 42 USD.
(a)
Find the value of mm[2 marks]
(b)
Find the value of cc and hence write down the equation of the model. [2 marks]
(c)
A rival company charges no booking fee but an hourly rate of 6.50 USD. Determine the number of hours at which both companies charge the same amount. [1 mark]
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2MasterySAQ-SEquation of a straight line and slope-intercept form7 marksPaper 1~11 min
A small business sells handmade candles. The weekly profit, PP dollars, is modelled as a linear function of the number of candles sold, xx. When 2020 candles are sold, the profit is USD 50\text{USD }50. When 5050 candles are sold, the profit is USD 230\text{USD }230.
(a)
Show that the equation of the line modelling the profit is P=6x70P = 6x - 70[3 marks]
(b)
Hence, find the minimum number of candles that must be sold each week for the business to break even. [2 marks]
(c)
The business owner claims the model predicts that selling 00 candles results in a loss of USD 70\text{USD }70. Evaluate whether this interpretation is meaningful in the context of the model. [2 marks]
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3ChallengeSAQ-LEquation of a straight line and slope-intercept form8 marksPaper 1~12 min
A small aircraft is flying at constant altitude on a straight path. At 14:00 its position is A(10,8,2)A(10, 8, 2), where coordinates are in kilometres relative to a control tower at origin OO, with xx East, yy North, and zz vertical. At 14:06 it passes point B(28,14,2)B(28, 14, 2).
(a)
Determine the equation of the flight path L1L_1 in parametric form, with t=0t = 0 at 14:00 and tt measured in minutes. [3 marks]
(b)
A second aircraft flies at a different altitude and follows the path L2:r=(12204)+s(420)L_2: \mathbf{r} = \begin{pmatrix}12\\20\\4\end{pmatrix} + s\begin{pmatrix}4\\-2\\0\end{pmatrix} where ss is measured in minutes after 14:00. Determine the time, to the nearest second, at which the two aircraft are closest, and find the minimum distance between them. [3 marks]
(c)
Analyse whether a collision between the two aircraft is possible. Justify your answer with reference to your results from parts (a) and (b). [2 marks]
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4FoundationSAQ-SEquation of a straight line and slope-intercept form5 marksPaper 1~8 min
The graph below shows a straight line passing through the points P(0,3)P(0, 3) and Q(4,11)Q(4, 11).
(a)
State the yy-intercept of the line. [1 mark]
(b)
Find the gradient of the line. [2 marks]
(c)
A second line has equation y=12x+ky = -\dfrac{1}{2}x + k, where kk is a constant. Determine the value of kk such that the two lines intersect at x=2x = 2[2 marks]
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5FoundationSAQ-SDefinitions of sine, cosine and tangent using right-angled triangles5 marksPaper 1~8 min
A ladder leans against a vertical wall. The base of the ladder is 1.5m1.5\,\text{m} from the wall, and the ladder makes angle of 72°72° with the horizontal ground.
(a)
Show that the height, hh, the ladder reaches up the wall satisfies h=1.5tan72°h = 1.5\tan 72°[2 marks]
(b)
Calculate the length of the ladder. Give your answer correct to 3 significant figures. [2 marks]
(c)
The base of the ladder is pushed 0.3m0.3\,\text{m} further from the wall. Determine the new angle the ladder makes with the ground, correct to 3 significant figures. [1 mark]
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6MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles5 marksPaper 1~8 min
A surveyor measures the height of a vertical cliff. From point AA on level ground, the angle of elevation to the top of the cliff is 3535^\circ. The surveyor walks 50m50\,\text{m} directly towards the cliff to point BB, from which the angle of elevation to the top of the cliff is 5555^\circ. Let hh metres be the height of the cliff and xx metres be the horizontal distance from BB to the base of the cliff.
(a)
Show that h=xtan55h = x\tan 55^\circ[1 mark]
(b)
Show that h=(x+50)tan35h = (x + 50)\tan 35^\circ[1 mark]
(c)
Find the value of xx, giving your answer correct to 3 significant figures. [2 marks]
(d)
Hence find the height of the cliff, giving your answer correct to 3 significant figures. [1 mark]
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7ChallengeSAQ-LDefinitions of sine, cosine and tangent using right-angled triangles8 marksPaper 1~12 min
A surveyor measures the height of a vertical cliff face. From point AA on level ground, the angle of elevation to the top of the cliff CC is 35°35°. The surveyor walks 50m50\,\text{m} directly towards the cliff to point BB. From BB, the angle of elevation to CC is 52°52°. Point DD is the base of the cliff. Let hh metres be the height of the cliff and xx metres be the horizontal distance BDBD.
(a)
Using triangle BCDBCD, write down an expression for hh in terms of xx[1 mark]
(b)
Using triangle ACDACD and your result from part (a), show that x=50tan35°tan52°tan35°x = \frac{50\tan 35°}{\tan 52° - \tan 35°} [3 marks]
(c)
Hence calculate the height of the cliff hh, giving your answer correct to 3 significant figures. [2 marks]
(d)
The surveyor's measuring instrument is held at eye level, 1.6m1.6\,\text{m} above the ground. Determine the percentage error introduced by ignoring this height when calculating hh[2 marks]
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8FoundationSAQ-SDefinitions of sine, cosine and tangent using right-angled triangles5 marksPaper 1~8 min
A surveyor stands on level ground and measures the angle of elevation to the top of a flagpole as 35°35° from a point PP, which is 20m20\,\text{m} from the base of the flagpole.
(a)
Show that the height of the flagpole is h=20tan35°h = 20\tan 35°[1 mark]
(b)
Calculate the height of the flagpole. Give your answer correct to 3 significant figures. [2 marks]
(c)
A second surveyor stands at point QQ, which is 8m8\,\text{m} further from the base of the flagpole than point PP, on the same level ground. Determine the angle of elevation to the top of the flagpole from point QQ, giving your answer correct to 1 decimal place. [2 marks]
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9FoundationSAQ-SLaw of Sines and its applications7 marksPaper 1~11 min
A surveyor needs to find the distance across a small lake. She identifies two points, AA and BB, on opposite sides of the lake. From a third point CC on the shore, she measures AC=120mAC = 120\,\text{m}, BC=95mBC = 95\,\text{m}, and AC^B=52°A\hat{C}B = 52°.
(a)
On the below, label the sides opposite to vertices AA, BB, and CC as aa, bb, and cc respectively, and write down the Law of Cosines in the form c2=c^2 = \ldots [1 mark]
(b)
Calculate the distance ABAB, giving your answer correct to 3 significant figures. [3 marks]
(c)
The surveyor claims that if the angle AC^BA\hat{C}B were increased to 90°90° while ACAC and BCBC remained unchanged, the distance ABAB would increase by more than 10m10\,\text{m}. Determine whether this claim is correct, justifying your answer. *(: triangle with vertices labelled AA, BB, CC; no side or angle values marked.)* [3 marks]
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10MasterySAQ-SLaw of Sines and its applications8 marksPaper 1~12 min
In triangle ABCABC, side a=8.0cma = 8.0\,\text{cm}, side b=10.0cmb = 10.0\,\text{cm}, and angle A=35°A = 35°.
(a)
Show that sinB=0.717\sin B = 0.717, correct to three significant figures. [3 marks]
(b)
Hence write down the two possible values of angle BB, correct to the nearest degree. [1 mark]
(c)
Determine which value(s) of BB give a valid triangle. For each valid triangle, find the corresponding length of side cc, correct to one decimal place. [4 marks]
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11ChallengeSAQ-LLaw of Sines and its applications8 marksPaper 1~12 min
A surveyor is mapping a triangular parcel of land ABCABC. The distance AB=120mAB = 120\,\text{m}, angle A^=52°\hat{A} = 52°, angle B^=63°\hat{B} = 63°, and BC=95mBC = 95\,\text{m}.
(a)
Determine the length of ACAC, correct to the nearest metre. [4 marks]
(b)
Point DD lies on BCBC such that ADAD bisects angle A^\hat{A}. Using the angle bisector theorem and the cosine rule, determine the length of ADAD, correct to 3 significant figures. [4 marks]
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12FoundationSAQ-SLaw of Sines and its applications5 marksPaper 1~8 min
In triangle PQRPQR, PQ=8.2cmPQ = 8.2\,\text{cm}, PR=6.5cmPR = 6.5\,\text{cm}, and QPR=40°\angle QPR = 40°.
(a)
Calculate the area of triangle PQRPQR, correct to 3 significant figures. [2 marks]
(b)
Calculate the length of side QRQR, correct to 3 significant figures. [3 marks]
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13MasterySAQ-SEquation of a straight line and slope-intercept form6 marksPaper 2~9 min
A company manufactures solar panels. The total cost, CC dollars, of producing xx panels is modelled by a linear function. When 200200 panels are produced, the total cost is USD 34000. When 500500 panels are produced, the total cost is USD 76000.
(a)
Calculate the slope of the linear cost function and interpret its meaning in context. [2 marks]
(b)
Determine the equation of the cost function in the form C=mx+cC = mx + c, where mm and cc are constants. [2 marks]
(c)
Each panel is sold for USD 180. Determine the minimum number of panels the company must sell to make a profit, and justify your answer. [2 marks]
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14MasterySAQ-SEquation of a straight line and slope-intercept form6 marksPaper 2~9 min
The graph below shows the temperature TT (in degrees Celsius) of a chemical solution as it cools over time tt (in minutes). The line passes through the points (5,72)(5, 72) and (20,42)(20, 42).
(a)
Calculate the gradient of this line. [2 marks]
(b)
Write down the meaning of the gradient in this context. [1 mark]
(c)
Determine the equation of the line in the form T=mt+cT = mt + c[2 marks]
(d)
The solution is considered safe to handle when its temperature falls below 30C30\,^\circ\text{C}. A technician claims the solution will be safe to handle after exactly 25 minutes. Determine whether the technician's claim is correct, justifying your answer. [1 mark]
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15MasterySAQ-SEquation of a straight line and slope-intercept form8 marksPaper 2~12 min
A small business sells handmade candles. The monthly profit, PP dollars, is modelled as a linear function of the number of candles sold, xx. When 5050 candles are sold, the profit is USD 200\text{USD }200. When 200200 candles are sold, the profit is USD 1250\text{USD }1250.
(a)
Determine the equation of the line in the form P=mx+cP = mx + c, stating the values of mm and cc[3 marks]
(b)
Interpret the value of mm and the value of cc in context. [2 marks]
(c)
The business owner claims that selling 180180 candles per month is sufficient to achieve a monthly profit of at least USD 1100\text{USD }1100. Determine whether this claim is correct, justifying your answer. [3 marks]
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16MasterySAQ-SDistance formula, midpoint formula and area of triangle8 marksPaper 2~12 min
A triangular park has vertices at points P(3,2)P(-3,\,2), Q(5,8)Q(5,\,8), and R(2,4)R(2,\,-4) on a coordinate grid where one unit represents 1010 metres.
(a)
Calculate the length of side PQPQ in metres. [2 marks]
(b)
Find the coordinates of the midpoint MM of side QRQR[1 mark]
(c)
Calculate the area of triangle PQRPQR in square units. [2 marks]
(d)
A straight path is built from PP through MM and extended until it meets side PRPR at point NN. Find the coordinates of NN and hence determine the ratio PN:NRPN : NR[3 marks]
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17MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles9 marksPaper 2~14 min
A surveyor measures the height of a vertical cliff face. From a point PP on level ground, the angle of elevation to the top of the cliff CC is 3232^\circ. The surveyor walks 50m50\,\text{m} directly towards the cliff to a point QQ. From QQ, the angle of elevation to CC is 4848^\circ. Let hh metres be the height of the cliff and xx metres be the horizontal distance from QQ to the base of the cliff BB.
(a)
Write down an expression for hh in terms of xx and tan48\tan 48^\circ[1 mark]
(b)
Using your answer to part (a) and the angle of elevation from PP, form an equation in xx alone and hence show that x=50tan32tan48tan32x = \frac{50\tan 32^\circ}{\tan 48^\circ - \tan 32^\circ} [3 marks]
(c)
Calculate the height of the cliff hh, correct to 3 significant figures. [1 mark]
(d)
The surveyor estimates that each angle of elevation could have a measurement error of up to ±1\pm 1^\circ. Determine the maximum possible value of hh under this error, and hence evaluate whether the measurement error has a significant effect on the calculated height. [4 marks]
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18MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles6 marksPaper 2~9 min
A ladder of length 6m6\,\text{m} leans against a vertical wall. The foot of the ladder rests on horizontal ground. The angle the ladder makes with the ground is θ\theta, where θ\theta is measured in degrees.
(a)
Write down expressions for the height hmh\,\text{m} of the top of the ladder above the ground, and the distance dmd\,\text{m} of the foot of the ladder from the wall, each in terms of θ\theta[2 marks]
(b)
Given that h=2dh = 2d, find the value of θ\theta[3 marks]
(c)
A safety regulation states that for a ladder of this length, the foot must be placed at least 1.5m1.5\,\text{m} from the wall. Determine whether the angle found in part (b) satisfies this regulation. Justify your answer. [1 mark]
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19MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles6 marksPaper 2~9 min
A kite is flying on a string. The string is straight and makes angle of 5555^\circ with the horizontal ground. The kite is at a vertical height of 40m40\,\text{m} above the ground.
(a)
Calculate the length of the string. Give your answer correct to 3 significant figures. [2 marks]
(b)
The wind changes and the kite moves so that the string now makes angle of 4040^\circ with the horizontal. The length of the string remains the same. (i) Determine the new vertical height of the kite above the ground. Give your answer correct to 3 significant figures. [2 marks]
(ii) Determine how much further the kite is horizontally from its original position. Give your answer correct to 3 significant figures. [2 marks]
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20MasterySAQ-SDefinitions of sine, cosine and tangent using right-angled triangles6 marksPaper 2~9 min
A ramp for loading vehicles inclined at 12°12° to the horizontal. The ramp has a vertical rise of 0.8m0.8\,\text{m}.
(a)
Calculate the length of the ramp, correct to 3 significant figures. [2 marks]
(b)
Safety regulations require that the angle of inclination does not exceed 8° for the same vertical rise of 0.8m0.8\,\text{m}. (i) Determine the minimum length of the new ramp that satisfies this regulation, correct to 3 significant figures. [2 marks]
(ii) The new ramp must fit within a loading bay of horizontal length 5.70m5.70\,\text{m}. Determine whether the new ramp fits within the loading bay. [2 marks]
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21MasterySAQ-SLaw of Sines and its applications6 marksPaper 2~9 min
A surveyor determines the width of a river. She stands at point AA one bank and observes a marker post at point BB directly opposite on the far bank, so that BAC=90°\angle BAC = 90°. She walks 50m50\,\text{m} along the bank to point CC. From CC, she measures BCA=62°\angle BCA = 62°. A second marker post at point DD is also on the far bank. From CC, the angle DCA=78°\angle DCA = 78°. The distance CD=35mCD = 35\,\text{m}.
(a)
Find the distance ABAB[3 marks]
(b)
Determine the distance ADAD. Give all answers correct to 3 significant figures. [3 marks]
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22MasterySAQ-SLaw of Sines and its applications6 marksPaper 2~9 min
A triangular plot of land has sides of lengths 120m120\,\text{m}, 150m150\,\text{m}, and 200m200\,\text{m}.
(a)
Calculate the area of the triangular plot. [3 marks]
(b)
A farmer wants to build a fence from one vertex perpendicular to the opposite side. Determine the length of the shortest such fence, justifying your answer. Give all answers correct to 33 significant figures. [3 marks]
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23MasterySAQ-SLaw of Sines and its applications6 marksPaper 2~9 min
A ship leaves port PP and sails 25km25\,\text{km} on a bearing of 040°040° to point QQ. It then changes course and sails 40km40\,\text{km} on a bearing of 130°130° to point RR.
(a)
Show that angle PQR=90°PQR = 90°, and hence calculate the distance PRPR[3 marks]
(b)
Determine the bearing of RR from PP. Give all answers correct to 3 significant figures. [3 marks]
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24MasterySAQ-SLaw of Sines and its applications6 marksPaper 2~9 min
A radio tower is located at point TT on horizontal ground. From point AA, which is 80m80\,\text{m} from the base of the tower on level ground, the angle of elevation to the top of the tower is 35°35°. From point BB, also on level ground but on the opposite side of the tower from AA, the angle of elevation to the top of the tower is 28°28°. Points AA, TT, and BB are collinear.
(a)
Calculate the height of the tower. [2 marks]
(b)
Determine the distance ABAB. Give all answers correct to 33 significant figures. [4 marks]
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25ChallengeLAQEquation of a straight line and slope-intercept form10 marksPaper 3~15 min
A telecommunications company is planning to install a fibre-optic cable connecting two towns, AA and BB. The cable must pass through a relay station at a point PP on a straight road running between two substations. Town AA is at (2,5)(2, 5) and town BB is at (10,11)(10, 11), with distances in kilometres. The road passes through substation XX at (0,2)(0, 2) and substation YY at (8,6)(8, 6).
(a)
Show that the equation of the road XYXY is y=12x+2y = \dfrac{1}{2}x + 2[3 marks]
(b)
The cable consists of two straight segments, from AA to PP and from PP to BB, where PP has xx-coordinate xx. Show that the total cable length LL, in kilometres, satisfies L=(x2)2+(12x3)2+(x10)2+(12x5)2.L = \sqrt{(x-2)^2 + \left(\tfrac{1}{2}x - 3\right)^2} + \sqrt{(x-10)^2 + \left(\tfrac{1}{2}x - 5\right)^2}. [3 marks]
(c)
Using your GDC, find the minimum value of LL and the coordinates of PP that achieve this minimum. [2 marks]
(d)
The reflection of A(2,5)A(2,5) across the line y=12x+2y = \dfrac{1}{2}x + 2 is the point A(6,1)A'(6, 1). Explain, using a geometric argument, why the minimum total cable length equals the straight-line distance ABA'B, and verify this numerically. [2 marks]
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26ChallengeLAQEquation of a straight line and slope-intercept form10 marksPaper 3~15 min
A logistics company uses a straight conveyor belt to move parcels between two sorting stations. The belt is modelled as a line segment. Station SS is at (2,7)(-2,\,7) and station TT is at (4,5)(4,\,-5) (distances in metres). A sensor is to be installed at a point QQ on the conveyor belt such that the sum of the squares of the distances from QQ to SS and TT is minimised.
(a)
Prove that the equation of the line representing the conveyor belt STST is y=2x+3y = -2x + 3[3 marks]
(b)
Let the coordinates of QQ be (x,2x+3)(x,\,-2x+3). Show that the sum of squared distances S=QS2+QT2\mathcal{S} = QS^2 + QT^2 can be expressed as S=10x220x+100\mathcal{S} = 10x^2 - 20x + 100[3 marks]
(c)
Find the coordinates of QQ that minimise S\mathcal{S}, and show that QQ is the midpoint MM of STST[2 marks]
(d)
The perpendicular bisector of STST passes through MM. Using this property, explain why MM minimises S\mathcal{S} for any point on line STST, and deduce what happens to S\mathcal{S} as QQ moves away from MM along the belt. [2 marks]
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27ChallengeLAQUnit circle and angle measurement10 marksPaper 3~15 min
A particle moves on a unit circle centred at the origin OO. Its position at time tt seconds is P ⁣(cosθ(t),sinθ(t))P\!\left(\cos\theta(t),\,\sin\theta(t)\right), where θ(t)\theta(t) is measured in radians from the positive xx-axis. The angular velocity is ω(t)=dθdt\omega(t)=\dfrac{d\theta}{dt}. The motion is defined by θ(t)=arcsin(t)\theta(t)=\arcsin(t) for 0t10\leq t\leq 1.
(a)
Write down the velocity vector P˙(t)=ddt ⁣(cosθ,sinθ)\dot{P}(t)=\dfrac{d}{dt}\!\left(\cos\theta,\,\sin\theta\right) in terms of ω(t)\omega(t) and θ(t)\theta(t), and hence show that the speed of the particle is P˙=ω(t)— \dot{P} — = — \omega(t) — for all tt in the domain. [3 marks]
(b)
Show that cos(θ(t))=1t2\cos(\theta(t))=\sqrt{1-t^2} for 0t10\leq t\leq 1, and hence prove that ω(t)=11t2\omega(t)=\dfrac{1}{\sqrt{1-t^2}}[4 marks]
(c)
Using your result from (b), evaluate 01/2ω(t)dt\displaystyle\int_{0}^{1/2}\omega(t)\,dt and verify that this equals θ ⁣(12)θ(0)\theta\!\left(\tfrac{1}{2}\right)-\theta(0). Hence interpret, in terms of arc length on the unit circle, what this integral represents. [3 marks]
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28ChallengeLAQUnit circle and angle measurement10 marksPaper 3~15 min
A sensor arm rotates about a fixed point OO. The tip of the arm traces a path modelled by a point on the unit circle. At time tt seconds (t0t \geq 0), the angle the arm makes with the positive xx-axis θ(t)=π3sin(2t)\theta(t) = \dfrac{\pi}{3}\sin(2t) radians.
(a)
(i) State the coordinates of the tip of the arm at time tt in terms of θ(t)\theta(t). [1]
(ii) Show that the coordinates of the tip always satisfy x2+y2=1x^2 + y^2 = 1[2 marks]
(b)
(i) Show that the angular velocity is ω(t)=2π3cos(2t)\omega(t) = \dfrac{2\pi}{3}\cos(2t). [2]
(ii) Calculate the maximum angular speed of the arm, justifying your answer using properties of cos(2t)\cos(2t)[2 marks]
(c)
The tip of the arm can only reach points on the unit circle corresponding to angles within the range of θ(t)\theta(t). - Determine the range of θ(t)\theta(t). - Hence deduce the arc of the unit circle that the tip can never reach, giving your answer as a set of angles θ\theta that are inaccessible. - Verify that the point (1,0)(-1, 0) lies in this inaccessible arc. [3 marks]
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29ChallengeLAQLaw of Sines and its applications10 marksPaper 3~15 min
A surveyor is mapping a triangular plot of land ABCABC with AB=120mAB = 120\,\text{m}, BC=150mBC = 150\,\text{m}, and CA=90mCA = 90\,\text{m}.
(a)
Using the Law of Cosines, show that cosB=0.8\cos B = 0.8 and hence find angle BB, giving your answer correct to one decimal place. [3 marks]
(b)
A second surveyor, working from a different reference, models a triangle PQRPQR in which PQ=120mPQ = 120\,\text{m}, QR=150mQR = 150\,\text{m}, angle Q=36.9°Q = 36.9°, and the side opposite angle PP is PR=dmPR = d\,\text{m}. Using the Law of Sines applied to triangle ABCABC from part (a), with the known angle BB and its opposite side CA=90mCA = 90\,\text{m}, find the two possible values of angle AA, correct to one decimal place. [4 marks]
(c)
Justify why the Law of Sines produces two candidate values for angle AA, and evaluate which value is geometrically valid for triangle ABCABC. Support your reasoning with reference to the angle sum of a triangle and the relative lengths of the sides. Clarification of labelling for (b): In triangle ABCABC, angle B36.9°B \approx 36.9° is opposite side CA=90mCA = 90\,\text{m}. Side BC=150mBC = 150\,\text{m} is opposite angle AA. Apply the Law of Sines to find angle AA[3 marks]
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30ChallengeLAQLaw of Sines and its applications10 marksPaper 3~15 min
A telecommunications tower stands vertically at point TT on a hillside. Two observation points AA and BB lie on level horizontal ground, with AB=200mAB = 200\,\text{m}. In the horizontal plane, the angle TAB=48°\angle TAB = 48° and the angle TBA=62°\angle TBA = 62°. The slant distance AT=180mAT = 180\,\text{m}. The angle of elevation of the top of the tower from point AA is 32°32°.
(a)
Show that the distance BTBT is 152m152\,\text{m}, correct to 3 significant figures. [4 marks]
(b)
Calculate hh, the vertical height of the top of the tower above the level ground, correct to the nearest metre. [3 marks]
(c)
Determine the angle of elevation of the top of the tower from point BB. Hence justify whether the angle of elevation from BB is greater than, equal to, or less than that from AA, explaining your reasoning in terms of the relative distances ATAT and BTBT[3 marks]
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