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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.
 

 Jun 9, 2017
 #1
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(Page is suppost to be side)

 Jun 9, 2017
 #2
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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         ?

 Jun 10, 2017

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