1 / 15

US5251

Solving Problems Modelled by Triangles. US5251. PYTHAGORAS. Can only occur in a right angled triangle. hypotenuse. Pythagoras Theorem states:. h. a. h 2 = a 2 + b 2. b. right angle. e.g. x. 9.4 cm. 7.65 m. y. 11.3 m. 8.6 cm. x 2 = 7.65 2 + 11.3 2. y 2 + 8.6 2 = 9.4 2.

raina
Download Presentation

US5251

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Solving Problems Modelled by Triangles US5251

  2. PYTHAGORAS Can only occur in a right angled triangle hypotenuse Pythagoras Theorem states: h a h2 = a2 + b2 b right angle e.g. x 9.4 cm 7.65 m y 11.3 m 8.6 cm x2 = 7.652 + 11.32 y2 + 8.62 = 9.42 9.42 = y2 + 8.62 square root undoes squaring smaller sides should always be smaller than the hypotenuse x2 = 186.2125 - 8.62 y2 = 9.42 – 8.62 - 8.62 x = √186.2125 y2 = 14.4 x = 13.65 m (2 d.p.) y = √14.4 y = 3.79 cm (2 d.p.)

  3. TRIGONOMETRY (SIN, COS & TAN) - Label the triangle as follows, according to the angle being used. to remember the trig ratios use SOH CAH TOA Hypotenuse (H) Opposite (O) and the triangles A Always make sure your calculator is set to degrees!! Adjacent (A) O A O S H C H T A means divide 1. Calculating Sides means multiply e.g. O 7.65 m x H h T A O O 29° 50° h = tan50 x 6.5 6.5 cm A O x = sin29 x 7.65 h = 7.75 cm (2 d.p.) x = 3.71 m (2 d.p.) S H

  4. e.g. d = 455 ÷ sin32 O d H 455 m d = 858.62 m (2 d.p.) O S H 32° 2. Calculating Angles -Same method as when calculating sides, except we use inverse trig ratios. e.g. B 4.07 m 23.4 mm H sin-1 undoes sin 2.15 m 16.1 mm H A O A cosB = 2.15 ÷ 4.07 sinA = 16.1 ÷ 23.4 A O B = cos-1(2.15 ÷ 4.07) A = sin-1(16.1 ÷ 23.4) C H S H B = 58.1° (1 d.p.) A = 43.5° (1 d.p.) Don’t forget brackets, and fractions can also be used

  5. TRIGONOMETRY APPLICATIONS e.g. A ladder 4.7 m long is leaning against a wall. The angle between the wall and ladder is 27°. Draw a diagram and find the height the ladder extends up the wall. A Wall (x) C H 27° A Ladder (4.7 m) x = cos27 x 4.7 H x = 4.19 m (2 d.p.) A e.g. A vertical mast is held by a 48 m long wire. The wire is attached to a point 32 m up the mast. Draw a diagram and find the angle the wire makes with the mast. C H H A 48 m cosA = 32 ÷ 48 32 m A = cos-1(32 ÷ 48) A A = 48.2° (1 d.p.)

  6. NON-RIGHT ANGLED TRIANGLES 1. Naming Non-right Angled Triangles - Capital letters are used to represent angles - Lower case letters are used to represent sides e.g. Label the following triangle c A The side opposite the angle is given the same letter as the angle but in lower case. B b C a

  7. 2. Sine Rule a = b = c . SinA SinB SinC a) Calculating Sides Only 2 parts of the rule are needed to calculate the answer e.g. Calculate the length of side p p = 6 . Sin52 Sin46 52° A × Sin52 p = 6 × Sin52 Sin46 × Sin52 B 46° 6 m b p = 6.57 m (2 d.p.) p a To calculate you must have the angle opposite the unknown side. Re-label the triangle to help substitute info into the formula

  8. b) Calculating Angles For the statement: 1 = 3 is the reciprocal true? 2 6 Yes as 2 = 6 1 3 Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is on top and therefore easier to calculate. You must calculate Sin51 before dividing by 6 (cannot use fractions) SinA = SinB = SinC a b c a = b = c . SinA SinB SinC e.g. Calculate angle θ Sinθ = Sin51 7 6 θ A Sinθ = Sin51 × 7 6 × 7 × 7 B 51° 6 m b θ = sin-1( Sin51 × 7) 6 7 m a θ = 65.0° (1 d.p.) To calculate you must have the side opposite the unknown angle Re-label the triangle to help substitute info into the formula

  9. Sine Rule Applications e.g. A conveyor belt 22 m in length drops sand onto a cone-shaped heap. The sides of the cone measure 7 m and the cone’s sides make an angle of 32° with the ground. Calculate the angle that the belt makes with the ground (θ), and the diameter of the cone’s base (x). SinA = SinB = SinC a b c Conveyor belt : 22 m 116° A b Sinθ = Sin148 7 22 b 7 m 7 m a 32° B 32° 148° B A θ Sinθ = Sin148 × 7 22 × 7 × 7 x a a = b = c . SinA SinB SinC θ = sin-1( Sin148 × 7) 22 x = 7 . Sin116 Sin32 θ = 9.7° (1 d.p.) × Sin116 x = 7 × Sin116 Sin32 × Sin116 x = 11.87 m (2 d.p.)

  10. 3. Cosine Rule -Used to calculate the third side when two sides and the angle between them (included angle) are known. a2 = b2 + c2 – 2bcCosA a) Calculating Sides e.g. Calculate the length of side x x2 = 132 + 112 – 2×13×11×Cos37 13 m b x2 = 61.59 A 37° x = √61.59 x x = 7.85 m (2 d.p.) a 11 m c Remember to take square root of whole, not rounded answer Re-label the triangle to help substitute info into the formula

  11. b) Calculating Angles - Need to rearrange the formula for calculating sides CosA = b2 + c2 – a2 2bc Watch you follow the BEDMAS laws! e.g. Calculate the size of the largest angle P CosR = 132 + 172 – 242 2×13×17 24 m a Q CosR = -0.267 13 m b A R = cos-1(-0.267) 17 m c R = 105.5° (1 d.p.) R Re-label the triangle to help substitute info into the formula Remember to use whole number when taking inverse

  12. Cosine Rule Applications e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °. Calculate the distance from the tee to the hole (x) and the angle (θ) at which the golfer hit the ball away from the correct direction. b x a Hole Tee a2 = b2 + c2 – 2bcCosA A θ x2 = 1302 + 2452 – 2×130×245×Cos60 x2 = 45075 130 m A 245 m 60° x = √45075 b a c c Ball x = 212.31 m (2 d.p.) CosA = b2 + c2 – a2 2bc Cosθ = 212.312 + 2452 – 1302 2×212.31×245 Remember to use whole number from previous question! Cosθ = 0.848 θ = cos-1(0.848) θ = 32.0° (1 d.p.)

  13. 3D FIGURES - Pythagoras and Trigonometry can be used in 3D shapes e.g. Calculate the length of sides x and w and the angles CHE and GCH A B x2 = 52 + 62 x = √52 + 62 G F x = √61 x = 7.8 m (1 d.p.) w O Make sure you use whole answer for x in calculation w2 = 72 + 7.82 D C 7 m w = √72 + 7.82 A x 5 m w = √110 O w = 10.5 m (1 d.p.) H E 6 m A tanCHE = 5 ÷ 6 tanCHE = 7 ÷ 7.8 O O CHE = tan-1(5 ÷ 6) CHE = tan-1(7 ÷ 7.8) T A T A CHE = 39.8° (1 d.p.) CHE = 41.9° (1 d.p.)

  14. 4. Area of a triangle - can be found using trig when two sides and the angle between the sides (included angle) are known Area = ½abSinC e.g. Calculate the following area Area = ½×8×9×Sin39 9 m b C Area = 22.7 m2 (1 d.p.) 39° 52° 8 m a 89° Re-label the triangle to help substitute info into the formula Calculate size of missing angle using geometry (angles in triangle add to 180°)

  15. Area Applications e.g. A ball is hit a distance of 245 m on a golf hole. The distance from the ball to the hole is 130 m. The angle between the hole and tee (from the ball) is 60 °. Calculate the area contained in between the tee, hole and ball. Hole Tee Area = ½abSinC C Area = ½×130×245×Sin60 130 m 245 m 60° a b Area = 13791.5 m2 (1 d.p.) Ball

More Related