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Trigonometry 17 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

@ganeshie8

OpenStudy (anonymous):

by only using sin law buddy

OpenStudy (campbell_st):

well I think it looks like this |dw:1415262618968:dw| 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...

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