1. Welcome to Triangles and Solving them This is mainly for higher grade students but foundation students can use the section on Pythagoras’ theorem Start Finish

2. Let’s Get Started Right-Angled Triangles Non Right-Angled Triangles Back

3. Right-Angled Triangles Question has 2 sides and wants the other Question involves 3D shapes Back Question involves angles

4. Pythagoras’ Theorem Back This theorem connect the three sides of a right-angle triangle; there are many ways in which it is expressed but here we will use the sequence Square SquareAdd/SubtractSquare root If finding the longest side we use the add and if one of the shorter the subtract 12 cm 8 cm a In this example, we are finding the shorter side and so will be subtracting 12 2 - 8 2 = 144 – 64 = 80 and so the length a is √80 = 8.94 cm Test questions

5. Pythagoras’ Examples 12 cm 13 cm x 10 cm x Back Solutions

6. Pythagoras’ Solutions 12 cm x 13 cm 10 cm x13 cm 13 2 + 12 2 = 169 + 144 = 313 x is √313 = 17.7 cm 13 2 - 10 2 = 169 - 100 = 69 x is √69 = 8.31 cm Back

7. SOHCAHTOA a Hypotenuse Adjacent Opposite HypSin Opp Cover up then gives the formula such as Hyp = Opp/Sin questions Back

8. Objective Can I use SOHCAHTOA with angles in right-angled triangles To Answer Back to menu 13cm 5cm x 7cm 13° x Q3. A ship leaves a port and sails 20 km east and then 30 km south. What bearing is the ship from the port Q1 Q2 30km 20km

9. Key points SOCAHTOA : Sin = opp/hyp; Cos= adj/hyp Tan = opp/adj Use the -1 to find and angle, button to find length Q1: Opp =5; hyp =13  Sin -1 (5/13) = 22.6° Q2: angle = 13°, adj = 7 ; hyp = 7 ÷ Cos 13° = 7.18cm Q3: opp = 30; adj = 20 : angle = Tan -1 (3/2)= 56.3° Bearing is 90 + 56.3° = 146.3 To Question Back to menu

10. Non-Right Angle triangles This needs one of three formulas depending on what you know 1. Three Sides(SSS) or 2 sides and the angle between(SAS) uses the Cosine rule a 2 = b 2 + c 2 – 2bcCosA 2. Any other combination uses the Sine Rule a=b=c Sin A Sin B Sin C can be written the other way up if angle is being found 3. Area of triangle = ½ abSinC (again requires SAS) Back to menu Sine rule Cosine ruleArea

11. COSINE RULE Three Sides(SSS) or 2 sides and the angle between(SAS) uses the Cosine rule a 2 = b 2 + c 2 – 2bcCosA Can be rewritten as Cos A = b 2 + c 2 –a 2 2bcnot on formula sheet Example 8cm 9cm 82° Find the other side (call it a) a 2 = 9 2 + 8 2 – 2x9x8xcos82° = 81 + 64 – 144Cos82° = 145 – 20 = 125 a =  125 = 11.2cm If we wanted one of the angles, we would now use the sine rule Find an angle Back to menu

12. Objective Can I find sides and angles in non right-angled triangles To Answer Back to menu Q3. A ship leaves a port and sails 20 km on a bearing of 070 and then 30 km on a bearing 120. How far from the port is it and on what bearing is the ship from the port Q1 – find the largest angle in a triangle of sides 7cm,8cm and 9cm Q2A C B Angle A = 50°, Angle B = 70° and AC = 12 cm. Find the length of AB

13. Key points Cosine Rule : a 2 = b 2 + c 2 – 2bcCosA (use for SSS and SAS) Sine Rule: a = b = calso area of Δ = ½ ab SinC (angle SinA SinB SinCbetween) Q1: cosine rule: largest angle opposite largest side 9 2 = 7² + 8² - 2x7x8xCosA gives 81= 113-112CosA (not CosA) Move 113 to give -32=-112CosA so CosA = (-32÷-112) so A=73.4° Q2: Use Sine rule but note C is 60° so AB ÷ Sin60= 12÷ sin 70 so that AB = 12 x Sin 60 ÷ Sin 70 = 11.1 cm Q3. from diagram To Question Back to menu P R Q T From question angle PQT=70 and RQT=60 making PQR=130. Using cosine rule PR² = 20² + 30² -20x30xcos130 = 1686 so PR=41.1km Using Sine Rule Sin P = 30x Sin 130 ÷ 41.1 so P= 34° so the bearing of the ship is 104° 70° 60° 70°

14. Objective Can I solve 3D Trigonometry Questions To Answer Back to menu ABCD is a square of side 7 cm and X is the midpoint of ABCD. M is the midpoint of AD and E is directly above X. Find a.Length EX b.Angle EMX c.Angle ECX

15. Key points Solve all problems by finding 2D triangles and solving them usually using Pythagoras and SOHCAHTOA a. The first step in finding EX is to find AC using the right-angled triangle ADC which will give AC as  (7² + 7²) = 9.90. From this AX = ½ of AC = 4.95. In Δ EAX we now know EA is 13 and AX = 4.95 so we can find EX using Pythagoras again, EX =  (13² - 4.95²) = 12.0 cm b.In Δ EMX for angle EMX, we now know that EX(Opp) is 12.0 and MX(Adj) is 3.5 ( ½ of 7) so that angle is tan -1 ( 12 ÷ 3.5) = 73.7° c.In Δ ECX for angle ECX, EC(Hyp) = 13 cm and CX(Adj) = 4.95 from part a. This gives us that ECX = Cos -1 ( 4.95 ÷ 13) = 67.6° To Question Back to menu

16. Area of Triangle The area of triangle can be found using ½ abSinC (again requires SAS that is the angle must be between the 2 sides) 8cm 82° 9 cm A = ½ ab SinC = ½ x 8 x 9 x Sin82 = 35.6 cm² Back to menu

17. Sine Rule a=b=c Sin A Sin B Sin C Usually written the other way up if finding an angle 8cm 82° 6 cm A CB As finding angle C Sin C = Sin A c a So Sin C = Sin A x c = sin 82 x 6 = 0.74 a 8 So C = sin -1 (0.74)= 48° Back to menu

18. Good luck with your exam Here are some MyMaths exercises to practice with Pythagoras SOHCAHTOA 3D trig Sine Rule Cosine Rule sides Cosine Rule angles Area of triangle Click anywhere to finish