We denote, by #hat(A,B)#, the angle btwn. the sides #A and B#.
#hat(A,B)=5pi/12; hat(B,C)=pi/2 rArrhat(C,A)=pi/12#
Now, in the #right/_^(ed)Delta# with #hat(B,C)=pi/2#, we have,
#tan(hat(A,B))=C/BrArrtan(5pi/12)=C/13rArrC=13(3.7321)=48.5173#
Therefore, the Area of the #right/_^(ed)Delta# [with #hat(B,C)=pi/2#],
#=1/2*B*C=1/2*13*48.5173=315.36245sq.unit#
But, wait a little! If you don't want to use the Table of Natural Tangents , see the Enjoyment of Maths below to find the value of #tan5pi/12# :-
#tan(5pi/12)=sin(5pi/12)/cos(5pi/12)=sin(5pi/12)/sin(pi/2-5pi/12)#
#=sin(5pi/12)/sin(pi/12)=sin(5theta)/sintheta, where, theta=pi/12#
#=(sin5theta-sin3theta+sin3theta-sintheta+sintheta)/sintheta#
#=(2cos4theta*sintheta+2cos2theta*sintheta+sintheta)/sintheta#
#={cancel(sintheta)(2cos4theta+2cos2theta+1)}/cancelsintheta#
#=2cos(4pi/12)+2cos2pi/12+1#
#=2*1/2+2*sqrt3/2+1#
#=2+sqrt3#
#2+1.7321#
#3.7321#
Isn't this Enjoyable Maths?!
