Presentation on theme: "Solving Problems Modelled by Triangles. PYTHAGORAS Can only occur in a right angled triangle Pythagoras Theorem states: hypotenuse right angle e.g. square."— Presentation transcript:
Solving Problems Modelled by Triangles
PYTHAGORAS Can only occur in a right angled triangle Pythagoras Theorem states: hypotenuse right angle e.g. square root undoes squaring smaller sides should always be smaller than the hypotenuse h 2 = a 2 + b 2 h a b x 7.65 m 11.3 m 9.4 cm y 8.6 cm x 2 = x 2 = x = √ x = m (2 d.p.) = y y = y 2 = – y 2 = 14.4 y = √14.4 y = 3.79 cm (2 d.p.)
TRIGONOMETRY (SIN, COS & TAN) - Label the triangle as follows, according to the angle being used. A Hypotenuse (H) Opposite (O) Adjacent (A) to remember the trig ratios use SOH CAH TOA and the triangles S O HC A HT O A means divide means multiply 1. Calculating Sides 29° e.g. x 7.65 m H O S O H x = sin29 x 7.65 x = 3.71 m (2 d.p.) 50° 6.5 cm h O A T O A h = tan50 x 6.5 h = 7.75 cm (2 d.p.) Always make sure your calculator is set to degrees!!
e.g. d 455 m 32° H OS O H d = 455 ÷ sin32 d = m (2 d.p.) 2. Calculating Angles -Same method as when calculating sides, except we use inverse trig ratios. A 16.1 mm 23.4 mm e.g. O H S O H sinA = 16.1 ÷ 23.4 sin -1 undoes sin A = sin -1 (16.1 ÷ 23.4) A = 43.5° (1 d.p.) Don’t forget brackets, and fractions can also be used B 2.15 m 4.07 m H A C A H cosB = 2.15 ÷ 4.07 B = cos -1 (2.15 ÷ 4.07) B = 58.1° (1 d.p.)
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. 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. Wall (x) Ladder (4.7 m) 27° H A C A H x = cos27 x 4.7 x = 4.19 m (2 d.p.) 48 m 32 m A H A C A H cosA = 32 ÷ 48 A = cos -1 (32 ÷ 48) A = 48.2° (1 d.p.)
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 a B C The side opposite the angle is given the same letter as the angle but in lower case. b c A
2. Sine Rule a = b = c. SinA SinB SinC a) Calculating Sides e.g. Calculate the length of side p p 6 m 52° 46° To calculate you must have the angle opposite the unknown side. Only 2 parts of the rule are needed to calculate the answer p = 6. Sin52 Sin46 × Sin52 p = 6 × Sin52 Sin46 p = 6.57 m (2 d.p.) Re-label the triangle to help substitute info into the formula A B a b
b) Calculating Angles For the statement: 1 = 3 is the reciprocal true? 2 6 Yes as 2 = Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is on top and therefore easier to calculate. a = b = c. SinA SinB SinC SinA = SinB = SinC a b c e.g. Calculate angle θ 7 m 6 m θ 51° Sinθ = Sin To calculate you must have the side opposite the unknown angle × 7 Sinθ = Sin51 × 7 6 θ = sin -1 ( Sin51 × 7) 6 θ = 65.0° (1 d.p.) You must calculate Sin51 before dividing by 6 (cannot use fractions) Re-label the triangle to help substitute info into the formula A B a b
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). Conveyor belt : 22 m θ 7 m 32° x 148° A a b B Sinθ = Sin × 7 Sinθ = Sin148 × 7 22 θ = sin -1 ( Sin148 × 7) 22 θ = 9.7° (1 d.p.) SinA = SinB = SinC a b c a = b = c. SinA SinB SinC 116°A a b B x = 7. Sin116 Sin32 × Sin116 x = 7 × Sin116 Sin32 x = m (2 d.p.)
3. Cosine Rule -Used to calculate the third side when two sides and the angle between them (included angle) are known. a 2 = b 2 + c 2 – 2bcCosA a) Calculating Sides e.g. Calculate the length of side x x 37° 13 m 11 m Re-label the triangle to help substitute info into the formula a A b c x 2 = – 2×13×11×Cos37 x 2 = x = √61.59 x = 7.85 m (2 d.p.) Remember to take square root of whole, not rounded answer
b) Calculating Angles - Need to rearrange the formula for calculating sides CosA = b 2 + c 2 – a 2 2bc e.g. Calculate the size of the largest angle P R Q 17 m 24 ma Ab c Re-label the triangle to help substitute info into the formula CosR = – ×13×17 Watch you follow the BEDMAS laws! CosR = Remember to use whole number when taking inverse R = cos -1 (-0.267) R = 105.5° (1 d.p.) 13 m
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. Hole 245 m Tee Ball 130 m x θ 60° a A bc a 2 = b 2 + c 2 – 2bcCosA x 2 = – 2×130×245×Cos60 x 2 = x = √45075 x = m (2 d.p.) CosA = b 2 + c 2 – a 2 2bc Cosθ = – ×212.31×245 Cosθ = θ = cos -1 (0.848) θ = 32.0° (1 d.p.) A a b c Remember to use whole number from previous question!
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 x w 6 m 7 m H GF E DC BA 5 m x 2 = x = √ x = √61 x = 7.8 m (1 d.p.) w 2 = w = √ w = √110 w = 10.5 m (1 d.p.) Make sure you use whole answer for x in calculation O A tanCHE = 5 ÷ 6 CHE = tan -1 (5 ÷ 6) CHE = 39.8° (1 d.p.) T O A O A T O A tanCHE = 7 ÷ 7.8 CHE = tan -1 (7 ÷ 7.8) CHE = 41.9° (1 d.p.)
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 52° 89° 8 m 9 m Re-label the triangle to help substitute info into the formula C a b 39° Calculate size of missing angle using geometry (angles in triangle add to 180°) Area = ½×8×9×Sin39 Area = 22.7 m 2 (1 d.p.)
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 245 m Tee Ball 130 m 60° a C b Area = ½×130×245×Sin60 Area = m 2 (1 d.p.) Area = ½abSinC