Question

one side of a parallelogram is 10'' long and make angles 45° and 75° with the diagonals. find the length of the other side

Answer 1

one side of a parallelogram is 10'' long and make angles 45° and 75° with the diagonals. find the length of the other side

Solution:

1)Angle (AOB) = 180° - (75°+45°) = 60°

2)Law of sines:

$$\cfrac{AB}{sin60°}=\cfrac{OB}{sin45°}$$ $$OB=\cfrac{AB*sin45°}{sin60°}=\cfrac{10*\cfrac{\sqrt{2}}{2}}{\cfrac{\sqrt{3}}{2}}=10\sqrt{\cfrac{2}{3}};$$

3)

$$BD=2*BO=20\sqrt{\frac{2}{3}}$$

4)Law of cosines:

$$AD^{2}=AB^{2}+BD^{2}-2*AB*BD*cos(\angle ABD);$$ $$AD^{2}=10^{2}+(20\sqrt{\cfrac{2}{3}})^{2}-2*10*20\sqrt{\cfrac{2}{3}}*cos75°;$$ $$AD^{2}=100+{\cfrac{800}{3}}-2*10*20\sqrt{\cfrac{2}{3}}*\cfrac{\sqrt{6}-\sqrt{2}}{4}$$ $$AD^{2}=\frac{500+200\sqrt{3}}{3}$$ $$AD=\frac{10\sqrt{15+6\sqrt{3}}}{3}$$

Answer:

$$AD=\frac{10\sqrt{15+6\sqrt{3}}}{3}$$