Part A, Multiple Choice (1 mark each, 25 marks total)
1In a right-angled triangle, which ratio defines the sine of an angle $\theta$?L1
A $\dfrac{\text{adjacent}}{\text{hypotenuse}}$
B $\dfrac{\text{opposite}}{\text{adjacent}}$
C $\dfrac{\text{opposite}}{\text{hypotenuse}}$
D $\dfrac{\text{hypotenuse}}{\text{opposite}}$
C, $\dfrac{\text{opposite}}{\text{hypotenuse}}$. SOH-CAH-TOA: S in = O pposite / H ypotenuse.
2In a right-angled triangle the reference angle is $32°$, the hypotenuse is $15\text{ cm}$, and $x$ is the side opposite the angle. Which equation correctly sets up the problem?L1
A $\sin 32° = \dfrac{x}{15}$
B $\cos 32° = \dfrac{x}{15}$
C $\tan 32° = \dfrac{x}{15}$
D $\sin 32° = \dfrac{15}{x}$
A, $\sin 32° = \dfrac{x}{15}$. The opposite ($x$) and hypotenuse ($15$) are involved, so use $\sin\theta = O/H$. Here $O = x$ and $H = 15$.
3The angle $34.7°$ is written in degrees and minutes. What is it, to the nearest minute?L1
A $34°7'$
B $35°42'$
C $34°70'$
D $34°42'$
D, $34°42'$. The whole part is $34°$. Multiply the decimal part by 60: $0.7 \times 60 = 42$ minutes. So $34.7° = 34°42'$.
4A right-angled triangle has reference angle $32°$, hypotenuse $15\text{ cm}$, and unknown opposite side $x$. Find $x$, correct to 2 decimal places. (Use $\sin 32° \approx 0.52992$.)L1
5To find an unknown angle when two sides are known, which calculator functions are required?L1
A $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$ (inverse trig)
B $\sin$, $\cos$, $\tan$ (direct trig)
C square root
D logarithm
A, inverse trig functions. When you know a ratio and want the angle, apply the inverse: if $\sin\theta = 0.6$ then $\theta = \sin^{-1}(0.6)$. Press SHIFT (or 2ND) then the trig key.
6In a right-angled triangle the reference angle is $48°$, the adjacent side is $9\text{ m}$, and the hypotenuse $H$ is unknown. Which equation correctly finds $H$?L2
A $H = 9 \times \cos 48°$
B $H = 9 \times \sin 48°$
C $H = \dfrac{9}{\cos 48°}$
D $H = \dfrac{9}{\tan 48°}$
C, $H = \dfrac{9}{\cos 48°}$. Adjacent and hypotenuse give $\cos 48° = \dfrac{9}{H}$. The unknown $H$ is in the denominator, so $H = \dfrac{9}{\cos 48°}$. The hypotenuse must be longer than $9\text{ m}$.
7A right-angled triangle has reference angle $48°$ and adjacent side $9\text{ m}$. Find the hypotenuse, correct to 2 decimal places. (Use $\cos 48° \approx 0.66913$.)L2
A $6.02\text{ m}$
B $13.45\text{ m}$
C $9.98\text{ m}$
D $8.10\text{ m}$
B, $13.45\text{ m}$. $\cos 48° = \dfrac{9}{H} \Rightarrow H = \dfrac{9}{\cos 48°} \approx \dfrac{9}{0.66913} \approx 13.45\text{ m}$. Option A wrongly multiplies and gives a hypotenuse shorter than the adjacent side.
8In a right-angled triangle the reference angle is $40°$ and the hypotenuse is $20\text{ cm}$. Find the side opposite the angle, correct to 2 decimal places. (Use $\sin 40° \approx 0.64279$.)L2
9A right-angled triangle has reference angle $25°$ and adjacent side $18\text{ cm}$. Find the hypotenuse, correct to 2 decimal places. (Use $\cos 25° \approx 0.90631$.)L2
A $19.86\text{ cm}$
B $16.31\text{ cm}$
C $8.39\text{ cm}$
D $42.60\text{ cm}$
A, $19.86\text{ cm}$. $\cos 25° = \dfrac{18}{H} \Rightarrow H = \dfrac{18}{\cos 25°} \approx \dfrac{18}{0.90631} \approx 19.86\text{ cm}$.
10In a right-angled triangle the side opposite $\theta$ is $9\text{ cm}$ and the adjacent side is $4\text{ cm}$. Find $\theta$ to the nearest minute.L3
13The angle of elevation from point A up to point B is $27°$. What is the angle of depression from B down to A?L4
A $63°$
B $153°$
C $27°$
D $54°$
C, $27°$. The angle of elevation from A to B equals the angle of depression from B to A. They are alternate interior angles between the two parallel horizontal lines, so both equal $27°$.
14A ramp is inclined at $15°$ to the horizontal and has a horizontal run of $12\text{ m}$. How high does the ramp rise, to 2 decimal places? (Use $\tan 15° \approx 0.26795$.)L4
A $3.22\text{ m}$
B $11.59\text{ m}$
C $44.78\text{ m}$
D $12.42\text{ m}$
A, $3.22\text{ m}$. Height $= d \cdot \tan\theta = 12 \times \tan 15° \approx 12 \times 0.26795 \approx 3.22\text{ m}$.
15An observer $80\text{ m}$ from the base of a cliff looks up at an angle of elevation of $32°$. How high is the cliff, to 2 decimal places? (Use $\tan 32° \approx 0.62487$.)L4
16From the top of a $45\text{ m}$ lighthouse, the angle of depression to a boat is $18°$. How far is the boat from the base of the lighthouse, to 1 decimal place? (Use $\tan 18° \approx 0.32492$.)L4
A $14.6\text{ m}$
B $138.5\text{ m}$
C $42.8\text{ m}$
D $47.3\text{ m}$
B, $138.5\text{ m}$. The depression from the top equals the elevation from the boat (alternate angles). $\tan 18° = \dfrac{45}{d} \Rightarrow d = \dfrac{45}{\tan 18°} \approx \dfrac{45}{0.32492} \approx 138.5\text{ m}$.
17A tree is $25\text{ m}$ from an observer, whose angle of elevation to the top of the tree is $52°$. Find the height of the tree, to 2 decimal places. (Use $\tan 52° \approx 1.27994$.)L4
18A ship sails on a true bearing of $070°$. What is the back bearing (the bearing back to the starting point)?L5
A $070°$
B $110°$
C $250°$
D $290°$
C, $250°$. Since the bearing is less than $180°$, add $180°$: back bearing $= 070° + 180° = 250°$.
19A bushwalker travels N$35°$E for $8\text{ km}$. How far north of her starting point is she, to 2 decimal places? (Use $\cos 35° \approx 0.81915$.)L5
A $4.59\text{ km}$
B $6.55\text{ km}$
C $9.76\text{ km}$
D $13.95\text{ km}$
B, $6.55\text{ km}$. The angle from the North line is $35°$. The North component $= d\cos\alpha = 8\cos 35° \approx 8 \times 0.81915 \approx 6.55\text{ km}$ north.
20Which true bearing is equivalent to the compass bearing S$70°$W?L5
A $070°$
B $110°$
C $200°$
D $250°$
D, $250°$. S$70°$W is in the SW quadrant. Measuring clockwise from North: $180° + 70° = 250°$T.
21A yacht sails $50\text{ km}$ on a true bearing of $200°$. How far west of its starting point is it, to 2 decimal places? (Use $\sin 20° \approx 0.34202$.)L5
A $32.14\text{ km}$
B $46.98\text{ km}$
C $17.10\text{ km}$
D $25.00\text{ km}$
C, $17.10\text{ km}$. $200°$T is in the SW quadrant. The angle from the South line is $200° - 180° = 20°$. The West (East-West) component $= d\sin\alpha = 50\sin 20° \approx 50 \times 0.34202 \approx 17.10\text{ km}$ west.
22Point C is on a true bearing of $210°$ from point A. What is the bearing from C back to A?L5
A $030°$
B $210°$
C $150°$
D $390°$
A, $030°$. Since $210° \geq 180°$, subtract $180°$: back bearing $= 210° - 180° = 030°$. The result must stay within $000°$ to $360°$.
23A ship leaves port and sails $20\text{ km}$ on bearing $040°$T to buoy B, then $15\text{ km}$ on bearing $130°$T to point C. Since the two bearings differ by $90°$, the legs are perpendicular. Find the straight-line distance PC.L5
A $35\text{ km}$
B $25\text{ km}$
C $5\text{ km}$
D $625\text{ km}$
B, $25\text{ km}$. The legs form a right angle, so by Pythagoras $PC = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\text{ km}$.
24Which true bearing is equivalent to the compass bearing N$25°$W?L5
A $025°$
B $205°$
C $155°$
D $335°$
D, $335°$. N$25°$W is in the NW quadrant. Measuring clockwise from North: $360° - 25° = 335°$T.
25A ship sails $85\text{ km}$ on a true bearing of $130°$. How far south of its starting point is it, to 2 decimal places? (Use $\cos 50° \approx 0.64279$.)L5
A $65.11\text{ km}$
B $54.64\text{ km}$
C $71.31\text{ km}$
D $42.50\text{ km}$
B, $54.64\text{ km}$. $130°$T is in the SE quadrant. The angle from the South line is $180° - 130° = 50°$. The South (North-South) component $= d\cos\alpha = 85\cos 50° \approx 85 \times 0.64279 \approx 54.64\text{ km}$ south.
Part B, Short Answer (show all working)
1L1
A right-angled triangle has a reference angle $\theta$, with the side opposite $\theta$ equal to $7\text{ cm}$ and the hypotenuse equal to $12\text{ cm}$.
(a) Write the trig equation that relates $\theta$ to these two sides.
(b) Find $\theta$ correct to 2 decimal places.
(c) Express $\theta$ in degrees and minutes, to the nearest minute.
(a) Opposite and hypotenuse are involved, so use sine: $\sin\theta = \dfrac{7}{12}$.
(c) The angle of depression is $40°$. It equals the angle of elevation because the two horizontal lines are parallel and the line of sight is a transversal, making them alternate interior angles.
5L4
From point A on level ground, the angle of elevation to the top T of a building is $42°$. From point B, which is $20\text{ m}$ closer to the building than A, the angle of elevation to T is $58°$.
(a) Let $d$ = horizontal distance from B to the building. Write two equations for $h$ (the height) in terms of $d$.
(b) Solve for $d$, then find $h$, correct to 1 decimal place. (Use $\tan 42° \approx 0.90040$, $\tan 58° \approx 1.60033$.)
(a) From A (distance $d + 20$ from building): $h = (d+20)\tan 42°$.
From B (distance $d$ from building): $h = d\tan 58°$.
(a) Convert the compass bearing S$40°$E to a true (three-figure) bearing.
(b) A plane flies from town P to town Q on a true bearing of $115°$. Find the bearing from Q back to P (the back bearing).
(c) Convert the true bearing $115°$ to a compass bearing.
(a) S$40°$E is in the SE quadrant. Measuring clockwise from North: $180° - 40° = 140°$T.
(b) $115° < 180°$, so add $180°$: back bearing $= 115° + 180° = 295°$T.
(c) $115°$ is in the SE quadrant ($090°$ to $180°$): compass bearing $= $ S$(180° - 115°)$E $= $ S$65°$E.
7L5
A ship leaves port P and sails $60\text{ km}$ on a true bearing of $030°$ to point Q. It then sails $40\text{ km}$ on a true bearing of $120°$ to point R.
(a) Find the eastward and northward displacements for the leg P to Q. (Use $\sin 30° = 0.5$, $\cos 30° \approx 0.86603$.)
(b) Find the eastward and northward displacements for the leg Q to R. (Note $120°$T is SE, so the angle from South is $60°$; the northward component is negative.)
(c) Find the straight-line distance PR, to 1 decimal place.
(a) P to Q (bearing 030°, angle from North $= 30°$):
A yacht leaves port P and sails $20\text{ km}$ on a true bearing of $040°$ to buoy B, then $15\text{ km}$ on a true bearing of $130°$ to point C.
(a) Explain why the legs PB and BC meet at a right angle, and find the straight-line distance PC.
(b) The bearing from P to C is $077°$T. Find the back bearing from C to P.
(a) The two bearings differ by $130° - 040° = 90°$, so the legs PB and BC are perpendicular. By Pythagoras: $PC = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25\text{ km}$.
(b) $077° < 180°$, so add $180°$: back bearing $= 077° + 180° = 257°$T.
Trigonometry Complete
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