I use the area of a triangle is 1/2absinC where a and b are two sides of the triangle and C is the angle between the two sides.
For an isosceles triangle you can use area= (b*h)/2 if you know the base angles of the triangle and the length of the base. It would be easier to understand if you draw an isosceles triangle ABC with base angles of 30 degrees and a base of 8cm. A line bisects the base (BC) at D from angle BAC so BD=BC=4cm. You end up with two right angles triangles ADC and ADB. You can work out AD using SOH CAH TOA as you know the angle is 30 and one side is 4cm. You then work out the area of the triangle by doing (AD*8)/2 .
Hope this helps!
Answered Sun 10th June, 2012 @ 10:25 by Dilly Edited by Dilly on Sun 10th June, 2012 @ 10:28
ohh yeahh dat equation ..duhhh.. im soo gonna flop my exam tmrw! -_-
you only use soh cah toa or Pythagoras theorem if it is a right angled triangle, if not you use sine rule or cosine rule or 1/2absinC! so for an isosceles you would use sine rule or cosine rule or 1/2absinC !