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?!
