Sunday, 23 April 2017

trigonometry - In an triangle the least angle is 45circ and the tangents of the angles are in A.P.If its area be 27 sq.cm.Find the lengths of its sides.



In an triangle the least angle is 45 and the tangents of the angles are in A.P.If its area be 27 sq.cm.Find the lengths of its sides.







Let A be the smallest angle.A=45.tanA,tanB,tanC are in arithmatic progression.

2tanB=tanA+tanC

2sinBcosB=sinBcosAcosC

cosAcosC=cosB2

12cosC=cosB2

cosBcosC=2

I am stuck here.Please help.


Answer



almagest has provided a good answer. This answer uses your idea.



Using that B+C=18045=135, you can have
cosBcosC=2cosB=2 cos(135B)=cosB+sinB,
i.e.2cosB=sinB

Squaring the both sides gives
4(1sin2B)=sin2BsinB=25,cosB=15
because B90, and so we have
sinC=sin(135B)=310



By the way,
27=12bcsinAbc=542
and so using the law of sines with (1)(2) gives
2RsinB2RsinC=542R=352where R is the circumradius.




Thus,
a=2RsinA=35,b=2RsinB=62,c=2RsinC=9.


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