Question

the bases of a trapezoid are 22 and 12 respectively. The angles at the extremities of one base are 65 degrees and 40 degrees respectively. Find the Two legs. Anyone can help me :D

Answer 1

it is Case I buddy ... i dnot know the figure and solution fo this can you help me?

Answer 2

@ganeshie8

Answer 3

by only using sin law buddy

Answer 4

well I think it looks like this



you need to find the value of y 1st... given the trapezoid has a height of h.

The dotted lines represent a rectangle with opposite sides of 12.

meaning the arms of the end triangles add to 10 units

so I started by saying

\[\tan(40) = \frac{h}{y}\]

and

\[\tan(65) = \frac{h}{10 - y}\]

making h the subject of both equations you get

\[h = ytan(40)~~~~and~~~~h = (10 - y)\tan(65)\]

you can equate the 2 equations since they both equal h

then

\[y \tan(40) = (10 - y) \tan(65)\]

so if you distribute and collect like terms you get

\[y[\tan(40 + \tan(65)] = 10 \tan(65)\]

which becomes

\[y = \frac{10\tan(65)}{\tan(40) + \tan(65)}\]

so you can now find the 2 shorter sides of the end triangles marked as

y and 10 - y

then its easy to find the missing sides... you need the cos ratios...