Pythagorean Theorem: Solving Trig Problems

1. Review Sheet Chapter Eight Pythagorean Theorem:a2 +b2 = c2where c is the longest side (hypotenuse) in the right triangle Common Pythagorean Triples: 3 – 4 – 5; 5 – 12 – 13; 8 – 15 – 17 and their multiples Triangle Trigonometry: Formulas are now on the formula sheet, but knowing them still helps Draw picture of triangle described in a word problem to help out with setting up proper trig function • Label each side of the triangle as H for hypotenuse (opposite 90°)O for opposite the given angleA for side adjacent to given angle • Set up an equation using the trig function and the variableSOH – CAH – TOA: Sin (angle) = side opposite from the angle / hypotenuse Cos (angle) = side adjacent to the angle / hypotenuse Tan (angle) = side opposite from the angle / side adjacent to the angle • Solve for the variable All triangular trig problems are solved in one of three ways: Test Taking Tips: Remember virtual alligator when comparing sides and their opposite angles (big side opposite big ) Tan 45° = 1; bigger the angle the larger the sine and the smaller the angle the larger the cosine Angle of Elevation or Angle of Depression Some Side Trig Fnc (angle) = ------------------------- Some other Side H O angle A

2. Sin and cos are always between 0 and 1; so first statement is false. Sin CAB and cos CBA are always equal since the opposite side of CAB is the adjacent side of cos CBA CAB  DAB only if ACBD is a square (y = x) Pythagorean theorem is true!! AD // CB because ACBD is a parallelogram

4. A ladder leans against a wall. The bottom of the ladder is 10 feet from the base of the wall, and the top of the ladder makes an angle of 25° with the wall.Find the length, x, of the ladder.

5. SSM: • AC must be greater than 4, but less than 8 (since any two sides must be greater than the third in a triangle). half of AC = 4 / sin 60° = 23

6. Ch 8 Coordinate Relations and Transformations • SSM: • x is opposite larger angle • eliminate A and B X is the adjacent side of the 20 angle. Use trig: cos 22 = x/80 80 cos 20 = x 75.17 = x

7. Ch 8 Coordinate Relations and Transformations • SSM: • right triangle  Pythagorean theorem (on formula sheet) Have to use each set of numbers in the Pythagorean theorem to see which works 202 + 212 = 292 400 + 441 = 841 841 = 841

8. Ch 8 Coordinate Relations and Transformations • SSM: • height is smaller than 14 • eliminate A and B Special case right triangles; side opposite 60 = ½ hypotenuse 3 hypotenuse = 14, so answer is ½(14) 3 = 7 3

9. Ch 8 Coordinate Relations and Transformations • SSM: • hypotenuse is largest side • length > 20 • Eliminate A and B 20 foot side is opposite of the 38 angle, so we can us sine sin 38 = 20 / hyp hyp sin 38 = 20 hyp = 20 / sin 38 hyp = 32.49

10. Ch 8 Triangles • SSM: • draw figure • diagonals bigger than the sides Diagonals of a square form a 45-45-90 degree triangle. Sides opposite a 45 angle is ½ hypotenuse 2. Diagonal is the hypotenuse so side is ½ (14) 2 = 7 2

11. Ch 8 Triangles • SSM: • Virtual Alligator • x across from bigger side so eliminate A and B No hypotenuse, so we must use tangent. tan x = 8/7.8 x = tan-1 (8/7.8) = 45.7

12. Ch 8 Triangles 20 cm 21 cm 29 cm • SSM: • hypotenuse is largest sideso either 21 or 29 Use the Pythagorean Theorem and the numbers to pick out which work. 202 + 212 = 292 400 + 441 = 841 841 = 841

13. Ch 8 Triangles • SSM: • across from smallest angle, so small side (eliminate A and B) x is opposite side from angle in right triangle. We use trig to find the answer. sin 22 = x/15 15 sin 22 = x 5.62=x