How is making a perpendicular bisector different to making an angle bisector
1 answer:
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Check the picture below.
now, notice, for the angle hAC, the hypotenuse is hA, and the adjacent side is CA, therefore,
![\bf cos(\theta)=\cfrac{adjacent}{hypotenuse}\qquad cos(hAC)=\cfrac{5}{hA}\implies hA=\cfrac{5}{cos(hAC)} \\\\\\ hA=\cfrac{5}{cos\left[ \frac{cos^{-1}\left( \frac{5}{13} \right)}{2} \right]}](https://tex.z-dn.net/?f=%5Cbf%20cos%28%5Ctheta%29%3D%5Ccfrac%7Badjacent%7D%7Bhypotenuse%7D%5Cqquad%20cos%28hAC%29%3D%5Ccfrac%7B5%7D%7BhA%7D%5Cimplies%20hA%3D%5Ccfrac%7B5%7D%7Bcos%28hAC%29%7D%0A%5C%5C%5C%5C%5C%5C%0AhA%3D%5Ccfrac%7B5%7D%7Bcos%5Cleft%5B%20%5Cfrac%7Bcos%5E%7B-1%7D%5Cleft%28%20%5Cfrac%7B5%7D%7B13%7D%20%5Cright%29%7D%7B2%7D%20%5Cright%5D%7D)
make sure your calculator is in Degree mode, if you need the angle in degrees.
now, notice, for the angle hAC, the hypotenuse is hA, and the adjacent side is CA, therefore,
make sure your calculator is in Degree mode, if you need the angle in degrees.
Answer:
Reduction; scale factor of .
Step-by-step explanation:
Hope this helps!
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