Trigonometry (Pythagoras and SOH CAH TOA) 12 Trigonometry (Pythagoras and SOH CAH TOA) 1 / 8 Part a) 10.649 cm 11.649 cm 12.649 cm This is a simple application of Pythagoras theorem. Ok, so we have to reverse it to find one of the "legs" but its simple enough! 2 / 8 Part b) 18.99 cm 16.99 cm 17.99 cm 3 / 8 Part bii) 65.38º 64.38º 66.38º Let's set this up as a SOH CAH TOA. I will draw a little guy at the angle I am trying to find, then label the sides with respect to him (and the angle): Now, as we have Opp and Hyp, we select the SOH: Plug in the values and solve: 4 / 8 Part a) 10^2 +(AC)^2 = 11^2 10^2 + 11^2=BC^2 (AB)^2 +10^2 = 11^2 I think the hardest part of this question is understanding what they are trying to say. To cover our backs let's state a general Pythagoras theorem, and then make it relevant to the question by substituting all the values that we can: 5 / 8 Part bi) 4.3826m 4.4826m 4.5826m Ok, so time to solve the setup from part a 6 / 8 Part a) the sum of the two shorter sides are equal to the length of the longest side the sum of the squares of the two shorter sides or a right angled triangle are equal to the square of the longest side 7 / 8 Part b) 233.8 (cm2) 232.8 (cm2) 231.8 (cm2) The area is given by finding the area of the red and the blue triangles. First, we need the height of the red triangle, let's use Pythagoras to solve this. the area of a triangle is (0.5)*base*height 8 / 8 Part a) 22.62º 20.62º 21.62º To make this problem easier, we will split it into two triangles. First, we need to work on the blue triangle to find the hypotenuse. This is a shared edge with the red triangle. Then we can work on the red triangle now we can use SOH CAH TOA to solve for the missing angle Your score is LinkedIn Facebook Twitter VKontakte