Right Angle Triangles Trigonometry Metric relations BA BH

Right Angle Triangles Trigonometry

Metric relations BA² = BH X BC A CA² = CH X CB AH² = HB X HC AB X AC = AH X BC C B H Example: from page 173 in your workbook, question 1 b) A Notice we are given the measurement of BC which is 9. You can solve for z by using Pythagoras. You can then solve for AH using AB X AC = AH X BC Then you can solve for x by using BA² = BH X BC Once you have x you can calculate 9 -x to find y. Answers z = 7. 2 y = 5. 76 x = 3. 24 h = 4. 32 z 5. 4 h B x y C H 9

Metric Relations (cont’d) To solve triangles using metric relations, you may have to use a system of equations. Example on page 173 # 1 a) B x 1. 5 H y 1. 2 A z C When looking at this question, we need to establish our “knowns” and “unknowns”. So far, we have measurements for AH and AB. By using our metric relations we can solve for x, y, and z. With only AB and AH, we do not have enough information to use only one formula. We have to use our “substitution method” to solve. How? Take the 3 rd formula and the first formula AH² = HB X HC BA² = BH X BC 1. 2² = xy 1. 5² = x (x + y) 1. 44 = xy 2. 25 = x² + xy Isolate y y = 1. 44 / x substitute for y 2. 25 = x²+x(1. 44/x) 2. 25 = x² + 1. 44 2. 25 – 1. 44 = x² x = 0. 9 Now that you have x you can solve for y by using the first formula: BA² = BH X BC 2. 25 = xy 2. 25 = (0. 9) (y) y = 1. 6 And finally to solve for z we can use Pythagoras for the bigger triangle or the fourth formula since we have 3 out of 4 variables: AB X AC = AH X BC (1. 5)(z) = (1. 2)(2. 5) z=2