This triangle will provide exact values for Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values 60º 2 60º 30º 2 3 1 This triangle will provide exact values for sin, cos and tan 30º and 60º

½ ½ 1 ∞ Exact Values x 0º 30º 45º 60º 90º 3 Sin xº Cos xº Tan xº 30º 2 30º 3 Exact Values x 0º 30º 45º 60º 90º Sin xº Cos xº Tan xº ½ 1 ∞ ½ 3

Exact Values 45º 2 1 1 45º 1 1 Consider the square with sides 1 unit We are now in a position to calculate exact values for sin, cos and tan of 45o

1 45º 2 Exact Values Tan xº Cos xº Sin xº 90º 60º 45º 30º 0º x 1 1

Naming the sides of a Triangle B a c C b A 26-Apr-17

2 sides and the angle in between (SAS) Key feature To find the area you need to know 2 sides and the angle in between (SAS) Area of ANY Triangle Remember: Sides use lower case letters of angles The area of ANY triangle can be found by the following formula. B a C Another version c b Another version A If you know C, a and b If you know A, b and c If you know B, a and c 26-Apr-17

Example : Find the area of the triangle. Area of ANY Triangle Example : Find the area of the triangle. B B The version we use is a 20cm c C C 30o 25cm A A b

Example : Find the area of the triangle. Area of ANY Triangle Example : Find the area of the triangle. E The version we use is d 10cm 60o f 8cm F e D

What Goes In The Box ? Key feature Remember (SAS) Calculate the areas of the triangles below: (1) 23o 15cm 12.6cm A =36.9cm2 (2) 71o 5.7m 6.2m A =16.7m2

Finding the Angle given the area 17 cm 95 cm2 12 cm P Q R Find the size of angle P p r Area = ½qrsinP 95 = ½ 17 × 12 sinP 95 = 102 sinP q sinP = 95 ÷ 102 = 0∙931 P = sin-1 0∙931 = 68∙6° 26-Apr-17

The Sine Rule The Rule The side opposite angle A is labelled a The side opposite angle C is labelled c The side opposite angle B is labelled b A B C c a b The Rule

Calculating Sides Using The Sine Rule P 34o 41o x 10m Q R Example 1 Find the length of x in this triangle. 10 sin 34° x sin 41° Now cross multiply.

Find the length of x in this triangle. 10m 133o 37o x D E F Example 2 Find the length of x in this triangle. 10 sin 37° x sin 133° Cross multiply = 12.14m

b c a Cross multiply Problem The balloon is anchored to the ground as shown in the diagram. Calculate the distance between the anchor points. A x sin 35° sin 75° 25 sin 70° 35° 25 m b c 75° 70° Cross multiply x m B C a

Find the unknown side in each of the triangles below: x = 6.7cm y = 21.8mm (1) 12cm 72o 32o x (2) 93o y 47o 16mm (3) 87o 89m 35o a (4) 143o g 12o 17m a = 51.12m g = 49.21m

Calculating Angles Using The Sine Rule Example 1. ao 45m 23o 38m Z Y X Find the angle ao 45 sin aº 38 sin 23º Cross multiply = 0.463 Use sin-1

Find the size of the angle bo Example 2. 143o 75m 38m bo Find the size of the angle bo = 0.305

Calculate the unknown angle in the following: (2) 14.7cm bo 14o 12.9cm (1) 14.5m 8.9m ao 100o bo = 16o ao = 37.2o (3) 93o 64mm co 49mm c = 49.9o

Sine Rule Basic Examples

In the examples which follow use the sine rule to find the required side or angle. B D 3.2 E 80° 5.8 C 5.7 A 74° 34° 4.6 A 76° C Find AB Find angle F F 9.6 Find angle B B

In the examples which follow use the sine rule to find the required side or angle. 8 B C 72° Find AB M 12 N 86° 7 T Find angle N 13.8 127° 36° P R Q Find PQ

A 9 B 75° 32° C Find BC R 60° S 78° 3.8 T Find RT I 27° 130° 4.1 G H Find GI 8 39° 10 Y X Z Find angle Z B 70° C 9 13 A Find angle A

Sine Rule (Bearings) Trigonometry

In the examples which follow use the sine rule to find the required side or angle. 1. Consider two radar stations Alpha and Beta. Alpha is 140 miles west of Beta. The bearing of an aero plane from Alpha is 032° and from Beta it is 316°. How far is the aeroplane from each of the radar stations?

Two ports Dundee and Stonehaven are 54 miles apart with Dundee approximately due south of Stonehaven. The bearing of a ship at sea from Stonehaven is 098° and from Dundee it is 048°. How far is the ship from Dundee? A ship is sailing north. At noon its bearing from a lighthouse is 240°. Five hours later the ship is 84km further north and its new bearing from the lighthouse is 290°. How far is the ship from the lighthouse now?

Two ports P and Q are 35 miles apart with P due east of Q Two ports P and Q are 35 miles apart with P due east of Q. The bearings of a ship from P and Q are 190° and 126° respectively. a) What is the bearing of port Q from the ship? b) How far is the ship from port P The bearing of an aeroplane from Aberdeen is 068° and from Dundee it is 030°. Dundee is 69 miles south of Aberdeen. How far is the aero plane from Aberdeen?

Two oil rigs are 80 miles apart with rig B being 80 miles east of rig A. The bearing of a ship from rig B is 194° and from A the bearing is 140°. Find the distance of the ship from rig A. At noon a ship lies on a bearing 070° from Dundee. The ship sails a distance of 45 miles south and at 3pm its new bearing from Dundee is 130°. How far is the ship from Dundee now? If the ship maintains the same speed, how long will it take to return to Dundee?

An aeroplane leaves Leuchers airport and flies 80 miles north. It then changes direction and flies 120 miles east. The plane now turns onto a bearing of 126° and flies a further 164 miles. Calculate the bearing and distance of the plane from Leuchers.

Question Solution 1 122.4 miles ; 103.8 miles 2 69.8 miles 3 95.0 kilometers 4 Bearing 306° ; 22.9 miles 5 56.0 miles 6 114.9 miles 7 48.8 miles ; 3 hours 15 minutes 8 236.2 miles ; bearing 269° Note: the last question requires the cosine rule as well as the sine rule.

The Cosine Rule can be used with ANY triangle as long as we have been given enough information. B a c C b A Given angle C Given angle B Given angle A 26-Apr-17

Identify sides a, b, c and angle Ao Write down the Cosine Rule for a Using The Cosine Rule Example 1 : Find the unknown side in the triangle below: x 5m 12m 43o A B C Identify sides a, b, c and angle Ao a = x b = 5 c = 12 Ao = 43o Write down the Cosine Rule for a a2 = b2 + c2 - 2bc cos A x2 = 52 + 122 - 2 x 5 x 12 cos 43o Substitute values x2 = 81.28 Square root to find “x”. x = 9.02m

Using The Cosine Rule 137o 17.5 m 12.2 m y P Q R Example 2 : Find the length of side QR Identify the sides and angle. p = y r = 12.2 q = 17.5 P = 137o Write down Cosine Rule for p p2 = q2 + r2 – 2pq cos P Substitute y2 = 12.22 + 17.52 – 2 x 12.2 x 17.5 x cos 137o y2 = 767.227 y = 27.7m

a2 = b2 + c2 – 2bc cosA p2 = 432 + 312 – 2 × 43 × 31 × cos78° Find the length of the unknown side in the triangles: b (1) 78o 43cm 31cm p A C a c B a2 = b2 + c2 – 2bc cosA p2 = 432 + 312 – 2 × 43 × 31 × cos78° p2 = 2255∙7 p = 47∙5 cm

Find the length of the unknown side in the triangles: (2) 8m 5∙2m 38o m m = 5∙05m 112º 17 mm 28 mm k k = 37∙8 mm (3)

Basic Examples 1. Finding a side. Cosine Rule Basic Examples 1. Finding a side. 5.6cm 39° C B 4.7cm A Find a C Find c 6cm B 32° A 1. 2. A Find b C Find a 2.7cm B 3. C 4. 2.4cm 7.4cm 3.8cm 36° 104° B A

5. 7. 6. 8. 10. 9. Find a. Find a 4.5cm 22° A C 1.7cm B 6cm A B 6.6cm 44° C B 8.7cm A Find c 67° 7. 6. 8cm P 4 R Q 134° Find q C M 8. 71° 3cm 5cm 10. N 40° 11.4cm T U 8.8cm Find v V Find m P 9.

Finding Angles Using The Cosine Rule The Cosine Rule formula can be rearranged to allow us to find the size of an angle This formula is cyclic, depending on the angle to be found

Finding Angles Using The Cosine Rule Example 1 : Calculate the unknown angle, xo . F d e Label and identify angles and sides E D f D = xo d = 11 e = 9 f = 16 e 2 + f 2 - d 2 cos D = Write the formula for cos D 2ef 92 + 162 - 112 cos x = Substitute values into the formula. 2 x 9 x 16 cos x = 0.75 Use cos-1 0∙75 to find x x = 41∙4o 38

Label and identify the sides and angle. Example 2: Find the unknown angle in the triangle: B Label and identify the sides and angle. B = yo b = 26 a = 13 c = 15 C A a 2 + c 2 - b 2 cos B = Write down the formula for cos B 2ac 132 + 152 - 262 cos y = Substitute values 2 x 13 x 15 cos y = - 0∙723 The negative tells you the angle is obtuse. Use cos-1 -0∙723 to find y y = 136.3o

Calculate the unknown angles in the triangles below: (1) 12.7cm 7.9cm 8.3cm C A ao ao = 37.3o B B (2) 10m 7m 5m bo bo =111.8o C D

Basic Examples 2. Finding an angle. 4.7cm C A A 3.6cm B 6cm 4.1 cm 1. 2. 5.6 cm C 3. 5cm 5.2cm 3.2cm Find C Find A 6cm Find B B C A A 10.5cm B C 1.9cm C 4cm 2.7 cm 4. C 3.6cm 5. 6. 5.8 cm 7.4 cm B 2.4cm 4.7cm A A Find B B Find A Find C

6.9cm Y L 6cm X B A 4.5cm 8. 9. 8.7cm 2.8 cm 7. M 6.6cm 7cm 8cm Find X 4.5cm Find the largest angle. Z K C Find L M 3cm 11.9cm V T 5cm N 11. 12. Q 8.8 cm 4.5cm 4 4 9.6 cm Find the smallest angle Q 10. U R P Find the largest angle 7.6cm Find all the angles

Cosine Rule or Sine Rule How to determine which rule to use Two questions 1. Do you know the length of ALL the sides? OR SAS 2. Do you know 2 sides and the angle in between? If YES to either of the questions then Cosine Rule Otherwise use the Sine Rule

Calculate the size of x in each of these diagrams

The Sine Rule Application Problems A D The angle of elevation of the top of a building measured from point A is 25o. At point D which is 15m closer to the building, the angle of elevation is 35o Calculate the height of the building. T B 35o 25o 10o 36.5 145o 15 m Angle TDA = 180 – 35 = 145o Angle DTA = 180 – 170 = 10o

The Sine Rule The angle of elevation of the top of a column measured from point A, is 20o. The angle of elevation of the top of the statue is 25o. Find the height of the statue when the measurements are taken 50 m from its base 180 – 110 = 70o 180 – 70 = 110o Angle ATC = 180 – 115 = 65o Angle BCA = Angle ACT = B T C A 25o 65o 110o 20o 70o =5.1 m 53.21 m 5o 50 m

The Cosine Rule Application Problems L N H B Bearing = 90 – 20 = 070° A fishing boat leaves a harbour (H) and travels due East for 40 miles to a marker buoy (B). At B the boat turns left and sails for 24 miles to a lighthouse (L). It then returns to harbour, a distance of 57 miles. Make a sketch of the journey. Find the bearing of the lighthouse from the harbour. (nearest degree) H 40 miles 24 miles B L 57 miles A Bearing = 90 – 20 = 070° N 20°

Not to Scale The Cosine Rule a2 = b2 + c2 – 2bcCosA P Q W An AWACS aircraft takes off from RAF Waddington (W) on a navigation exercise. It flies 530 miles North to a point (P) as shown, It then turns left and flies to a point (Q), 670 miles away. Finally it flies back to base, a distance of 520 miles. Find the bearing of Q from point P. P 670 miles W 530 miles Not to Scale Q 520 miles P = 48.7° (49°) Bearing = 180 + 49 = 229°

1. Find the length of the third side of triangle ABC when COSINE RULE 1. Find the length of the third side of triangle ABC when i) b = 2, c = 5,  A = 60° ii) a = 2, b = 5,  A = 65° iii) a = 2, c = 5,  B = 115° iv) b = 6, c = 8,  A = 50° Find QR in ∆PQR in which PR = 4, PQ = 3,  P = 18° Find XY in ∆ XYZ in which YZ = 25, XZ = 30,  Z = 162° In ∆ ABC, a = 7, b = 4,  C = 53°. Calculate c.

5. In ∆ ABC, b = 4·2, c = 6·5,  A = 24°. Calculate a. In ∆ ABC, a = 1·64, c = 1·64,  B = 110°. Calculate b. In ∆ ABC, a = 18·5, b = 22·6,  C = 72·3°. Calculate c. 8. In ∆ABC, a = 100, b = 120,  C = 15°. Calculate c. 9. In ∆ABC, b = 80, c = 100,  A = 123°. Calculate a.

10. A town B is 20 km due north of town A and a town C 10. A town B is 20 km due north of town A and a town C is 15 km north-west of A. Calculate the distance between B and C. 11. Two ships leave port together. One sails on a course of 045° at 9 km/h and the other on a course of 090° at 12 km/h. After 2h 30 min, how far apart will they be? 12. From a point O, the point P is 3 km distant on a bearing of 040° and the point Q is 5 km distant on a bearing of 123°. What is the distance between P and Q ?

COSINE RULE (1) Solutions 1. Find the length of the third side of triangle ABC when i) a2 = 22 + 52  225cos60° = 4 + 25  200·5 = 19  a = 4·36 ii) c2 = 22 + 52  225cos65° = 4 + 25  200·4226 = 20·54763  c = 4·533 iii) b2 = 22 + 52  225cos115° = 4 + 25  225(0·42262 ) = 29 + 8·45237 = 37·45237  b = 6·12 iv) a2 = 62 + 82  268cos50° = 36 + 64  960·64279 = 100  661·7076 = 38·29239  a = 6·188

QR2 = 42 + 32  243cos18° = 16 + 9  240·951057 = 2·17464  QR = 1·4747 XY2 = 252 + 302  22530cos162° = 625 + 900  1500(0·951057 ) = 1525 + 1426·5848 = 2951·5848  XY = 54·33 4. c2 = 72 + 42  274cos53° = 31·2984  c = 5·5945

5. a2 = 4·22 + 6·52  24·26·5cos24° = 17·64 + 42·25  554·60·91355 = 10·0104  a = 3·164 6. b2 = 1·642 + 1·642  21·641·64cos110° = 22·6896  22·6896(0·34202 ) = 7·2190  b = 2·6868 7. c2 = 18·52 + 22·62  218·522·6cos72·3° = 598·778  c = 24·50 8. c2 = 1002 + 1202  2100120cos15° = 1217·780  c = 34·90

9. a2 = 802 + 1002  280100cos123° = 25114·225  a = 158·47 North 10. BC2 = 152 + 202  21520cos45° = 225 + 400  6000·7071 = 200·7359 BC = 14·17 km B 20 km C 11. QR2 = 302 + 22·52  23022·5cos45° = 900 + 506·25  954·59 = 451·6558 QR = 21·25 km 15 km 45° West A North Q 22·5 km 12. PQ2 = 32 + 52  235cos83° = 9 + 25  300·12187 = 30·344 PQ = 5·51 km 45° R P 30 km

Cosine Rule Bearings problems

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 1. Ship1 [ 74km, 053° ] ; Ship2 [ 104km, 112° ]  Port Ship1 Ship2 North 112° 53° 59°

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 2. Ship1 [ 56km,021° ] ; Ship2 [ 66km,090° ]  Port Ship1 Ship2 North

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 3. Ship1 [ 83km,060° ] ; Ship2 [ 80km,159° ]  Port Ship1 Ship2 North

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 4. Ship1 [ 50km,090° ] ; Ship2 [ 62km,142° ]  Port Ship1 Ship2 North

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 5. Ship1 [ 80km,146° ] ; Ship2 [ 70km,190° ]  Port Ship1 Ship2 North

Cosine Rule- Bearings:-Two Ships. In each example the distances and bearings of two ships from a port are given. Use the cosine rule to find the distance between the two ships. 6. Ship1 [ 47km,180° ] ; Ship2 [ 54km,235° ]  Port Ship1 Ship2 North

Bearings Problems Dundee is 84 km due south of Aberdeen. A ship at sea is on a bearing of 078° from Aberdeen and 048° from Dundee. How far is the ship from Dundee and from Aberdeen ? An aeroplane is 240 km from an airport on a bearing of 100° while a helicopter is 165 km from the airport on a bearing of 212°. How far apart are the aircraft ?

Bearings Problems A ship sails 80 km on a bearing of 060° from its home port. It then sails 93 km on a bearing of 134°. How far is it now from its home port ? Glasgow airport is 73 km from Edinburgh airport and lies to the west of Edinburgh airport. The bearing of an aeroplane from Glasgow airport is 040° while its bearing from Edinburgh airport is 300°. How far is the aeroplane from each airport ?

Bearings Problems A ship sails 74 km south from its port to a lighthouse. It then sails 85 km on a bearing of 160°. How far is the ship from the port ? From a port P, ship A is 144 km distant on a bearing of 036° and ship B is 97 km distant on a bearing of 114°. What is the distance between the two ships ?

Bearings Problems A ship sails 93 km on a bearing 054° and then another 108 km on a bearing of 110°. How far is it now from its starting point ? 8. Two radar stations Alpha and Beta pick up signals from an incoming aircraft. Alpha is 40 km east of Beta and picks up the signals on a bearing of 300°. Beta picks up the signals on a bearing of 070°. How far is the aircraft from Alpha and from Beta ?

aeroplane Q is 170 km from the same airport on a bearing of 094°. Bearings Problems 9. Aeroplane P is 200 km from an airport on a bearing of 208° while aeroplane Q is 170 km from the same airport on a bearing of 094°. How far apart are the two aeroplanes ? 10. The bearings and distances from Aberdeen of three oil platforms are : a) 028° 116 km b) 081° 104 km c) 138° 97 km A supply boat leaves Aberdeen, visits each platform and then returns to Aberdeen. Find the total length of the journey.