Trig help

[sgt_corpse]

Find the lenths of the diagonals of a parallelogram having sides of the given lengths and an angle of the given measures

6 cm, 10 cm, 60 degrees

The parallelegram is shaped like a kite our teacher did not tell us where to put the numbers please help.

 

[Gene]

It doesn't matter. No matter how you draw it the diagonals are the same. I pick the 10's horizontally with the 60° lower left. Starting there, going clockwise it is ABCD.
Under C, to AD extended is E.
The height is the 3sqrt(3)
Triangle CDE is a 3-6-3sqrt(3)
That makes AE=13 and ACE a right triangle with legs of 13 & 3sqrt(3).
BD is the hypot of another right triangle with legs of 7 & 3sqrt(3).

 

[soroban]

Hello, sgt_corpse!

Find the lengths of the diagonals of a parallelogram having sides 6 cm and 10 cm, and an angle of 60 degrees.

            A                 B
            * - - - - - - - - *
           /                 /
          /                 /
        6/                 /
        /                 /
       / 60°             /
      * - - - - - - - - *
      D        10       C

Since this posted under "Trig", I assume you know the Law of Cosines.


Draw diagonals AC and BD.


\(\displaystyle AC^2\:=\:6^2\,+\,10^2\,-\,2\cdot6\cdot10\cdot\cos60^o\:=\:76\)

. . Hence: .\(\displaystyle AC\:=\:\sqrt{76}\:\approx\:8.72\)


\(\displaystyle BD^2\:=\:6^2\,+\,10^2\,-\,2\cdot6\cdot10\cdot\cos120^o\:= \:196\)

. . Hence: .\(\displaystyle BD\:=\:\sqrt{196}\:=\:14\)