A composite geometric shape of triangle and rectangle is given with the proportionate relation given in the figure. Find angle #alpha# and #theta#?

enter image source here

1 Answer
May 2, 2016

#theta = arctan(2/3) ~~ 33.69^@#

#alpha= 90^@ - arctan(1/3)-arctan(1/4)~~57.53^@#

Explanation:

Labeling the picture as follows (with the additional constructed segment #bar(AE)#):


From the right triangle #triangleBCF#, given #bar(CF)=b/2=h# and #bar(BF) = (2h)/3#, we have #tan(theta) = ((2h)/3)/h=2/3#

Applying the inverse tangent function gives us #theta#:

#theta = arctan(2/3)~~33.69^@#


Note that #bar(EC) = bar(EF)-bar(CF) = bar(AD)-bar(CF)=(2b)/3-b/2 = b/6 = h/3#

Together with #bar(AE) = h#, we have the angle #angleEAC = arctan((h/3)/h) = arctan(1/3)#

Additionally, as #bar(BD) = h/3# and #bar(AD) = (2b)/3 = (4h)/3#, we have the angle #angleDAB = arctan((h/3)/((4h)/3)) = arctan(1/4)#

Noting that #angleEAD# is a right angle, we can now obtain #alpha#:

#alpha = angleEAD - angleEAC - angleDAB#

#= 90^@ - arctan(1/3)-arctan(1/4)~~57.53^@#