Help me please allright (member, Triangle)

cardboard adolescent · 06-10-2006, 05:05 PM

well you have to draw a line parallel to ED from B, and then where that intersects EA you have a new point, H. triangle AHB has side AH with length 1, hypotenuse AB with length 7, and side BH with length sqrt(48). you can then use the law of sines to say that sin90/7 = sinx/sqrt(48) and then solve for x. that gives you the angle measure of EAC. since you know angle E is 90, angle C is just 90 - <EAC.

06-10-2006, 05:05 PM	#21 (permalink)
cardboard adolescent ;) Join Date: Nov 2005 Location: CA Posts: 3,511	well you have to draw a line parallel to ED from B, and then where that intersects EA you have a new point, H. triangle AHB has side AH with length 1, hypotenuse AB with length 7, and side BH with length sqrt(48). you can then use the law of sines to say that sin90/7 = sinx/sqrt(48) and then solve for x. that gives you the angle measure of EAC. since you know angle E is 90, angle C is just 90 - <EAC.