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
1√2cosC=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=180∘−45∘=135∘, you can have
cosBcosC=√2⇒cosB=√2 cos(135∘−B)=−cosB+sinB,
i.e.2cosB=sinB
Squaring the both sides gives
4(1−sin2B)=sin2B⇒sinB=2√5,cosB=1√5
because B≤90∘, and so we have
sinC=sin(135∘−B)=3√10
By the way,
27=12bcsinA⇒bc=54√2
and so using the law of sines with (1)(2) gives
2RsinB⋅2RsinC=54√2⇒R=3√5√2where R is the circumradius.
Thus,
a=2RsinA=3√5,b=2RsinB=6√2,c=2RsinC=9.
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