A triangle ABC is such that sides a,b,c are in G.P. and sinA,sinB,sinC are in A.P., then the △ABC is
(where a,b,c are sides (in units) of △ABC opposite to ∠A,∠B and ∠C respectively)
A
Equilateral triangle
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B
Right angled triangle
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C
scalene triangle
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D
None of these
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Solution
The correct option is A Equilateral triangle Given : Sides a,b,c are in G.P. ⇒b2=ac⋯(i)
and sinA,sinB,sinC are in A.P., ⇒2sinB=sinA+sinC
Using sine rule : asinA=bsinB=csinC=2R ⇒2b=a+c⇒4b2=a2+c2+2ac⇒4ac=a2+c2+2ac[from(i)]⇒(a−c)2=0⇒a=c ∴a=b=c[from(i)]