In the figure; AC = BC and BC = 0.5AD
A) Calculate the angle A in the triangle BAD
B) Make the appropriate length for the AC in meter page (choose an appropriate size on page AC) and determine the area of the BAD triangle. Please provide different suggestions on how the area of the triangle BAD can be determined.
+9466 I got a strange answer for part A) , but I will post it anyway....
∠BAD = ∠BAC - ∠DAC
tan(∠BAC) = BC / AC Since AC = BC , replace " AC " with " BC ".
tan(∠BAC) = BC / BC
tan(∠BAC) = 1
∠BAC = arctan(1)
∠BAC = 45º
cos(∠DAC) = AC / AD
BC = 0.5AD Multiply both sides of the equation by 2 .
2BC = AD Since AC = BC , replace " BC " with " AC ".
2AC = AD
cos(∠DAC) = AC / AD Replace " AD " with " 2AC " .
cos(∠DAC) = AC / (2AC) Reduce fraction by AC
cos(∠DAC) = 1/2
∠DAC = arccos(1/2)
∠DAC = 60º
∠BAD = ∠BAC - ∠DAC
∠BAD = 45º - 60º
∠BAD = -15º
......
Maybe it is supposed to be BD = 0.5AC or DC = 0.5AC ?
