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Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
Math project by Shehribane
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Math project by Shehribane

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Trigonometry Project

Trigonometry Project

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  • 1. MIND MAP
  • 2. SOHCAHTOA 1. You use Sin when you have the opposite andSOH = 0 Sin hypotenuse H 2. You use Cos when you have the adjacent and hypotenuseCAH = A Cos H 3. You use Tan when you have the opposite and the adjacentTOA = O Tan A
  • 3. Labeling the Right Triangle 1. The angle is either at the top or bottomOPP HY 2. The opposite is always P opposite of the angle 3. The Hypotenuse is always the longest side 4. Adjacent is the side ADJ that is not labelled yet
  • 4. Trigonometry: The branch of mathematics concerned with the properties of triangles and calculations based on theseproperties.Acute Triangle : A triangle which all interior angles are acute anglesObtuse Triangle : A triangle in which one of the angles is an obtuse angle that is an angle greater than 90 o and less than 180oRight Triangle: A triangle with one right anglePrimary Trig Ratio : The basic ratios of trigonometry sine, cosine, tangentSine : is one of the primary trig ratios in a right angle triangle for opposite over hypotenuseSine Law : in any of triangle ABC with sides abc opposite angles A,B,C respectivelyCosine : is one of the primary trig ratios in a right angle triangle for adjacent over hypotenuseCosine Law: an equation used when a triangle does not have a right angle all 3 sides or 2 sides angle between LawPythagorean theorem: in a right angle triangle the square of the length of the hypotenuse is equal to the sun of the square ofthe squares of the lengths of the other two sidesTangent: one of the primary trig ratios in a right angle triangle the ratio of the length of the side opposite angle to the length ofthe adjacent sideHypotenuse: In a right angle triangle the side opposite the right angleAngle of Elevation: An angle measured from the horizontal plane upward from an observers eye to a given point above theplaneAngle of Depression : The angle measured below the horizontal that an observer must look to see an object that is lower thanthe observerBearings: A bearing is an angle, measured clockwise from the north direction
  • 5. An airplane is approaching a runway at an angle of descent of 30o What is the altitude of the airplane when it is 15 km along its flight path from the runway H yp • Label The Triangle 15 • Decide which TrigOPP km Ratio will be used • Sin, Cos, Tan 30o Adj
  • 6. 1. Use Tan because where already have H Hypotenuse and weOPP yp want to find the 15km opposite 30o Tan 30o = O 2. Calculate Tan of 30 Adj A then cross multiply Tan 30o = x 3. Multiply 0.5773 15 by 15 4. Gives you 0.5773 = x 8.67cm 15 Therefore, The altitude is X=8.67cm 8.67com
  • 7. Louise is a naturalist studying the effect of acid rain on fish population in different lakes. As part of her research she needs to know the length of lake Lebarge. Louise makes the measurements shown. How Long is the Lake. A a = b = c Sin 320 x 340 = Sin 960c 96o Sin A Sin B Sin C 0.5299 x 340 = 0.99452c c b 180.17254 = 0.99452c 0.99452 0.99452 340 = b = c 520 32o 181. 2cm = cB 340 cm C Sin 96 Sin 52 Sin 32 a Therefore, Lake Lebarge is 340 = c 181.2 cm long. Sin 96 Sin 32 3. Multiply 0.5299 with 340 and 0.99452 with c 4. Divide both side by 0.99452 1. Fill in the correct information in the equation 2. Cross Multiply the 340 with sin of 32 and sin 96 with c
  • 8. Angle S = 560 Length a = 14 cm Length t = 25 cm A Farmer is building a fence around his chicken pen Find the perimeter of the Triangle SAT with the given lengths and angles 1. Fill in all the angles S2 = a2 + t2 – 2at Cos S and lengths 2. Do 14 squared and S S2 = 142 + 252 – 2(14) (25) Cos 56 25 squared then multiply -2 by 14 and 25 560 S2 = 196 + 625 – 700 Cos 56 3. Do Cos of 56o and S2 = 196 + 625 – 700 0.5591 multiply it by 700 25cm 14cm 4. Add 196 and 625 then S = 196 + 625 – 391.37 2 subtract 391.37 t a Square root 429.63 to give S = 429. 63 you 20.7 S= 20.7 5. Add all the sides together to give you the perimeter of 59.73 TA s P=s+s+s 20cm P= 25 + 14 + 20.7 Therefore, the perimeter of the P = 59.73 triangle is 59.73
  • 9. BY: ShehribaneHaziri

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