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Trigonometric Ratios (5.1 – 5.5)

5.1 Review of the trigonometric ratios, using the unit circle.

5.2 Trigonometric ratios of: - q, 90 - q, 180 ± q, 360 ± q

5.3 The exact ratio

5.4 Bearings and angles of elevation.

5.5 Sine and cosine rules for a triangle. Area of a triangle, given two sides and the included angle.

Scroll down to page 2.

Trigonometry 1

Q.1. Use your calculator to find the value of the following to 4 decimal places:

(i) sin 30^o (ii) cos 40^o (iii) tan 60^o (iv) sin 37^o (v) sin 83⁰

(vi) sin 16^o (vii) cos 36^o (viii) cos 68^o (ix) tan 43^o (x) tan 26^o

Q.2. Use your calculator to find the value of the following to 4 decimal places:

(i) sin 27^o30’ (ii) cos 53⁰20’ (iii) tan 42^o12’ (iv) sin 49^o15’ (v) sin 76⁰52’

(vi) sin 12^o13’ (vii) cos 37^o5’ (viii) tan 52^o8’ (ix) cos 58^o42’ (x) tan 17^o8’

Q.3. Find the value of q to the nearest degree for the following:

(i) sin q = 0.7071 (ii) cos q = 0.7760 (iii) tan q = 1.540

(iv) sin q = 0.8910 (v) tan q = 0.4040 (vi) tan q = 0.7265

(vii) cos q = 0.8763 (viii) sin q = 0.6211 (ix) cos q = 0.2807 (x) tan q = 17.343

Q.4. Find the value of q to the nearest minute for the following:

(i) sin q = 0.5835 (ii) cos q = 0.9145 (iii) tan q = 10.5789

(iv) sin q = 0.6794 (v) tan q = 0.3083 (vi) tan q = 1.3384

(vii) cos = 0.9853 (viii) sin q = 0.9735 (ix) cos q = 0.7799 (x) tan q = 1.3009

Q.5. The diagram below shows the equilateral triangle ABC. Each side is 2 units long. The line AD is drawn to meet the mid-point of BC at D.

(i) Show that the triangles ABD and ACD are congruent.

(ii) Show that BAD = CAD = 30^o

(iii) Use Pythagoras’ theorem to find the exact length of AD.

(iv) Without using your calculator, write down the values of (a) sin 30^o

(b) sin 60^o (c) cos 30^o (d) cos 60^o (e) tan 30^o (f) tan 60^o

Answers:

Q.1. (i) 0.5000 (ii) 0.7660 (iii) 1.7321 (iv) 0.6018 (v) 0.9925

(vi) 0.2756 (vii) 0.8090 (viii) 0.3746 (ix) 0.9235 (x) 0.4877

Q.2. (i) 0.4617 (ii) 0.5972 (iii) 0.9067 (iv) 0.7576 (v) 0.7779

(vi) 0.2116 (vii) 0.7978 (viii) 1.2861 (ix) 0.5195 (x) 0.3083

Q.3. (i) 45^o (ii) 39^o (iii) 57^o (iv) 63^o (v) 22^o (vi) 36^o (vii) 29⁰ (viii) 38^o (ix) 74^o (x) 87^o

Q.4. (i) 35^o42’ (ii)23^o52’ (iii) 84^o36’ (iv) 42⁰48’ (v) 17^o8’ (vi) 53^o14’ (vii) 9^o 51’ (viii) 76^o47’ (ix) 38⁰45’ (x) 52^o27’

Q.5. (i) Show by SSS (ii) BAC = 60^o (Equilateral D) Since Ds then BAD = CAD = 30⁰ (iii) units (iv) (a) (b) (c) (d) (e) (f)

TRIGONOMETRY 2

Determine the length of the sides and the value of the angles represented by pronumerials in the following diagrams:

Answers:

a = 60^o b = 8 units c = 6.93 units d = 40^o e = 3.2 units

f = 3.83 units g = 10 units h = 53^o8’ i = 36^o52’ j = 35^o

k = 4.59 units l = 6.55 units m = 55^o n = 2.29 units p = 3.28 units

q = 60^o r = 10 units s = 17.32 units t = 65^o u = 13.59 units

v = 6.34 units w = 13.42 units x = 48^o11’ y = 41^o49’ z = 27^o

aa = 19.82units ab = 17.66 units ac = 3 units ad = 2.05 units

ae = 3.21 units af = 18^o45’ ag = 9.47 units

TRIGONOMETRY 3

Q.1. ABC is an isosceles right-angled triangle with AB = BC = 1 unit.

Calculate the length AC by Pythagoras’ theorem and hence determine the exact values of; (i) sin 45^o (ii) cos 45^o (iii) tan 45^o

Q.2. ABC is an equilateral triangle with each side 2 units long. AD is drawn perpendicular to BC, meeting BC at D.

(a) Show DABD DACD and hence show BD = DC = 1 unit and BAD = CAD = 30^o.

(b) Determine the exact values of; (i) sin 60^o (ii) cos 60^o (iii) tan 60^o (iv) sin 30^o (v) cos 30^o (vi) tan 30^o

Q.3. ABC is a right angled triangle with AB = 5 units and BC = 12 units.

Determine the length AC by Pythagoras’ theorem and hence determine the exact values of; (i) sin ACB (ii) cosACB (iii) tanACB

(iv) sinBAC (v) cosBAC (vi) tanBAC

ANSWERS

Q.1. AC = (i) (ii) (iii) 1

Q.2. (a) AB = AC = 2 units (ABC is an equilateral triangle)

ABD = ACD = 60^o (ABC is an equilateral triangle)

ADB = ADC = 90^o (given that AD perpendicular to BC)

Hence DABD DACD (AAS)

Since BC = 2 units and BD = DC, BD & DC = 1 unit each.

Since BAC = 60^o and BAD = CAD then BAD & CAD are 30^o each

(b) (i) (ii) ½ (iii) (iv) ½ (v) (vi)

Q.3. AC = 13 units (i) (ii) (iii) (iv) (v) (vi)

TRIGONOMETRY: BEARINGS

Bearings are indicated as an angle measured clockwise from north and are given in standard 3-figure notation. The letter T is usually written after the angle to indicate the direction from True north. Hence the direction east is written as 090^oT.

Q.1. Write down the bearings of the following directions:

(i) South (ii) west (iii) south-west (iv) north-west

(v) 30^o east of north (vi) 20^o south of east (vii) 10^o north of west

(viii) 17^o south of west (ix) 17^o west of south (x) 34^o north of east

Q.2. A boat is 10 nautical miles from a lighthouse at a bearing 030^oT. How far is the boat (i) east of the lighthouse
(ii) north of the lighthouse?

Q.3. A boat is 20 nautical miles from a lighthouse at a bearing 240^oT. How far is the boat (i) west of the lighthouse
(ii) south of the lighthouse?

Q.4. A car and a utility leave the same town. The car is driven 150 km north while the utility is driven 200 km west. When they have reached their destination,

(i) how far is the car from the utility?

(ii) what is the bearing of the car from the utility? (Answer to nearest degree)

(iii) what is the bearing of the utility from the car? (Answer to nearest degree)

Q.5. A lighthouse keeper sees a launch at a bearing of 090^oT and a yacht at a bearing of 150^oT. If the launch is 15 nautical miles due north of the yacht,

(i) how far is the yacht from the lighthouse?

(ii) what is the bearing of the yacht from the launch?

(iii) what is the bearing of the launch from the yacht?

Q.6. A bandicoot and a goanna are released into the wild from the same place. The bandicoot walks at a bearing of 060^oT while the goanna walks at a bearing of 150^oT. When they have both walked the same distance from their release point they are 800m apart.

(i) how far has each of them walked? (to nearest metre)

(ii) what is the bearing of the bandicoot from the goanna? (to nearest degree)

(iii) what is the bearing of the goanna from the bandicoot? (to nearest degree)

Q.7. Tom the cat is chasing Jerry the mouse. If Jerry is at a bearing of 137^o from Tom, What is the bearing of Tom from Jerry?

Q.8. A plane flies at a bearing of 214^o at a speed of 400 km/h for 3 hours. How far is the plane (i) west of its starting point (ii) south of its starting point (to nearest km)

Q.9. A boat is 10 nautical miles due north of a lighthouse. If it sails at a bearing of 210^oT at 5 knots (1 knot = 1 nautical mile per hour) for 4 hours, what is its new bearing from the lighthouse?

Answers:

Q.1. (i) 180^oT, (ii) 270^oT, (iii) 225^oT, (iv) 315^oT, (v) 030^oT, (vi) 110^oT,

(vii) 280^oT, (viii) 253^oT, (ix) 197^oT, (x) 056^oT

Q.2. (i) 5 nautical miles (ii) 8.66 nautical miles

Q.3. (i) 17.32 nautical miles (ii) 10 nautical miles

Q.4. (i) 250 km (ii) 053^oT (iii) 233^oT

Q.5. (i) 17.32 nautical miles (ii) 180^oT (iii) 000^oT

Q.6. (i) 566m (ii) 015^oT (iii) 195^oT

Q.7. 317^oT

Q.8. (i) 671 km (ii) 995 km

Q.9. 234^oT

TRIGONOMETRY: ANGLES of ELEVATION & DEPRESSION; BEARINGS

Q.1. Kulsoom was standing on the top of a coastal cliff and looking at Laura who was in a boat below. The angle of depression of Laura from Kulsoom was 30^o.

The cliff was measured as being 100 metres high.

(i) What was the angle of elevation of Kulsoom from Laura?

(ii) Draw a diagram of the above situation and clearly label the angles of elevation and depression.

(iii) How far was Laura from the base of the cliff?

Q.2. Kulsoom was on the 100 metres high cliff and observed a buoy, 40 metres from the base of the cliff. What was the angle of depression of the buoy?

Q.3. Laura then observed Kulsoom at the top of a different cliff. Laura noted that the angle of elevation was 40^o. She rowed her boat 50 metres towards the cliff and noted that the angle of elevation was 60^o. What was the height of the cliff?

Q.4. Laura wanted to determine the height of a tree. She measured 50 metres from the base of the tree and used a sight tube at that point to determine the angle of elevation as 20^o. How tall was the tree?

Q.5. A ball is rolling in a direction 30^o south of east. What is the ball bearing?

Q.6. Laura and Kulsoom were in opposing teams for an orienteering exercise. Both started from the same point at the same time. Laura jogged for 20 minutes at 6.0 km/h at a bearing of 150^o. Kulsoom walked for 45 minutes at 4.0 km/h at a bearing of 240^o. Both then sat down to rest.

(i) How far apart were Laura and Kulsoom when they were resting?

(ii) What was the bearing of Kulsoom from Laura?

(iii) What was the bearing of Laura from Kulsoom?

ANSWERS

Q.1. (i) 30^o

(ii)

(iii) 173.2 metres

Q.2. 68^o 12’

Q.3. 81.38 metres

Q.4. 18.2 metres

Q.5. 120T^o

Q.6. (i) km = 3606metres

(ii) 273^o41’T

(iii) 093^o41’T

TRIGONOMETRY: SINE & COSINE RULE

Q.1. (i) Find the length of the sides and the values of the angles in the following triangles where such measurements are not already shown.

(ii) Find the area of each triangle.

Q.2. (i) Find the length of the sides and the values of the angles in the following triangles where such measurements are not already shown.

(ii) Find the area of each triangle.

Q.3. (i) Find the length of the sides and the values of the angles in the following triangles where such measurements are not already shown.

(ii) Find the area of each triangle.

Answers:

Q.1. (i) AB = 12 cm, BAC = 36^o52’, BCA = 53^o8’, Area = 54 cm²

(ii) AC = 9.22 cm, BAC = 49^o24’, BCA = 40^o36’, Area = 21 cm²

(iii) BC = 4.62 cm, AB = 9.24 cm, BCA = 60^o, Area = 18.48 cm²

Q.2. (i) ABC = 41^o49’, ACB = 108^o11’, AB = 11.4 cm, Area = 22.8 cm²

(ii) BAC = 48^o45’, ABC = 61^o15’, AC = 9.33 cm, Area = 35.07 cm²

(iii) BAC = 42^o05’, ABC = 87^o55’, AC = 10.44 cm, Area = 27.99 cm²

Q.3. (i) BAC = 38^o56’, ABC = 70^o32’, ACB = 70^o 32’, Area = 25.45 cm²

(ii) AC = 5.79 cm, ACB = 88^o57’, BAC = 51^o03’, Area = 20.25 cm²

(iii) AB = 13.76 cm, BAC = 33^o50’, ABC = 16^o10’, Area = 19.15 cm²

TRIGONOMETRY – ANGLES IN ALL QUADRANTS

Do not use your calculator for this worksheet.

Q.1. Write down the exact value of: (i) sin 30^o (ii) cos 30^o (iii) tan 30^o

(iv) sin 45^o (v) cos 45^o (vi) tan 45^o (vii) sin 60^o

(viii) cos 60^o (ix) tan 60^o

Q.2. Write down the exact value of: (i) cosec 30^o (ii) sec 30^o (iii) cot 30^o

(iv) cosec 45^o (v) sec 45^o (vi) cot 45^o (vii) cosec 60^o

(viii) sec 60^o (ix) cot 60^o

Q.3. Write down the exact value of: (i) sin 150^o (ii) cos 150^o (iii) tan 150^o

(iv) sin 135^o (v) sec 135^o (vi) cot 135^o (vii) cosec 120^o

(viii) cos 120^o (ix) cot 120^o (x) sec 150^o

Q.4. Write down the exact value of: (i) sin 210^o (ii) cos 240^o (iii) tan 210^o

(iv) sin 225^o (v) sec 225^o (vi) cot 225^o (vii) cosec 210^o

(viii) cosec 240^o (ix) cot 240^o (x) sec 210^o

Q.5. Write down the exact value of: (i) sin 300^o (ii) cos 330^o (iii) tan 330^o

(iv) sin 315^o (v) sec 315^o (vi) cot 315^o (vii) cosec 330^o

(viii) cosec 300^o (ix) cot 330^o (x) sec 300^o

Answers:

Q.1. (i) ½ (ii) (iii) (iv) (v) (vi) 1 (vii) (viii) ½ (ix)

Q.2. (i) 2 (ii) (iii) (iv) (v) (vi) 1

(vii) (viii) 2 (ix)

Q.3. (i) ½ (ii) - (iii) - (iv) (v) - (vi) -1

(vii) (viii) - ½ (ix) - (x) -

Q.4. (i) -½ (ii) -½ (iii) (iv) - (v) - (vi) 1

(vii) -2 (viii) - (ix) (x) -

Q.5. (i) - (ii) (iii) - (iv) - (v)

(vi) -1 (vii) -2 (viii) - (ix) - (x) 2

TRIGONOMETRY-RATIOS

Consider a circle of radius 1 unit with an angle q drawn in the first quadrant as shown.

Sin q = = = y Cos q = = = x

Tan q = = =

By Pythagoras’ theorem, x² + y² = 1 i.e sin²q + cos²q = 1

sin²q + cos²q = 1

Dividing through by sin²q gives: + =

1 + cot²q = cosec²q

Dividing through by cos²q gives: + =

tan²q + 1 = sec²q

Q.1. Prove the following identities:

(i) tanq cosq = sinq (ii) 4 – 3 sin²q = 3 cos²q + 1

(iii) sec²q sin²q = tan²q (iv) sin²q + tan²q = sec²q - cos²q

(v) cosec²q - sin²q = cos²q + cot²q (vi) cosq cotq = cosecq - sinq

(vii) cot²q sin²q + cos²q +2sin²q = 2 (viii) tanq + cotq = secq cosecq

(ix) cosec²q + sec²q = tan²q + cot²q + 2

(x) cosec²q + sec²q = 1 / (sin²q cos²q)

Q.2. Solve the following trigonometric equations for 0^oq360^o :

(i) sinq = ½ (ii) 3sinq = 4cosq

(iii) cosecq = 3sinq (iv) 5sinq = 3cosq

(v) 4cotq = tanq (vi) cosecq = -5

(vii) cos²q =1 (viii) tanq + 1 = 0

(ix) sinq = (x) secq = -

Answers: Q.2.

(i) 30^o, 150^o (ii) 53^o8’, 233^o8’ (iii) 35^o16’, 144^o44’

(iv) 30^o58’, 210^o58’ (v) 63^o26’, 116^o34’, 243^o26’, 296^o34’

(vi) 191^o32’, 348^o28’ (vii) 0^o, 180^o, 360^o, (viii) 135^o, 315^o

(ix) 45^o, 135^o (x) 135^o, 225^o