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Welcome to Triangles and Solving them

Welcome to Triangles and Solving them. This is mainly for higher grade students but foundation students can use the section on Pythagoras’ theorem. Finish. Start. Let’s Get Started. Right-Angled Triangles. Non Right-Angled Triangles. Back. Right-Angled Triangles.

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Welcome to Triangles and Solving them

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

  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 angles Question involves 3D shapes Back

  4. Pythagoras’ Theorem 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 Square Add/Subtract Square root If finding the longest side we use the add and if one of the shorter the subtract In this example, we are finding the shorter side and so will be subtracting 122 - 82 = 144 – 64 = 80 and so the length a is √80 = 8.94 cm 12 cm 8 cm a Test questions Back

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

  6. Pythagoras’ Solutions x 13 cm 13 cm x 12 cm 10 cm 132+ 122 = 169 + 144 = 313 x is √313 = 17.7 cm 132 - 102 = 169 - 100 = 69 x is √69 = 8.31 cm Back

  7. SOHCAHTOA In a right-angled triangle where an angle is involved, trigonometry is used and the mnemonic to remember it is SOHCAHTOA. Depending on which two sides are involved, we use one of three rations Sine = Cosine = and Tangent To use we find which two sides are involved and then remember that the button(e.g. sin) is used to find a length and the -1 (e.g. sin-1) for an angle Formula triangles may be used such as Hypotenuse Opposite a Cover up then gives the formula such as Hyp = Opp/Sin Adjacent Opp Back Sin Hyp questions

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

  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 Back to menu To Question

  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 a2 = b2 + c2 – 2bcCosA2. 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 Cosine rule Area Sine rule

  11. COSINE RULE Three Sides(SSS) or 2 sides and the angle between(SAS) uses the Cosine rule a2 = b2 + c2 – 2bcCosA Can be rewritten as Cos A = b2 + c2 –a2 2bc not on formula sheet Example Find the other side (call it a) a2 = 92 + 82 – 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 8cm 9cm 82° Back to menu Find an angle

  12. Objective Can I find sides and angles in non right-angled triangles Q1 – find the largest angle in a triangle of sides 7cm,8cm and 9cm 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 Q2 A B C Angle A = 50°, Angle B = 70° and AC = 12 cm. Find the length of AB Back to menu To Answer

  13. Key points Cosine Rule : a2 = b2 + c2 – 2bcCosA (use for SSS and SAS) Sine Rule: a = b = c also area of Δ = ½ ab SinC (angle SinA SinB SinC between) Q1: cosine rule: largest angle opposite largest side 92 = 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 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° Q 70° 70° 60° P T R Back to menu To Question

  14. Objective Can I solve 3D Trigonometry Questions • 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 • Length EX • Angle EMX • Angle ECX Back to menu To Answer

  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 • 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° • 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° Back to menu To Question

  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) A = ½ abSinC = ½ x 8 x 9 x Sin82 = 35.6 cm² 9 cm 8cm 82° 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 A 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° 6 cm 82° B C 8cm Back to menu

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

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