The angles A,B and C of a triangle ABC are in A.P.  and a:b=1:\sqrt3.

If c = 4 cm , then the area ( in sq. cm) of this triangle is: 

  • Option 1)

    \frac{2}{\sqrt3}

  • Option 2)

    4\sqrt3

  • Option 3)

    2\sqrt3

  • Option 4)

    \frac{4}{\sqrt3}

 

Answers (1)

In \Delta ABC ,

A,B,C are in A.P.

=> \angle B=60^{\circ}

Now, it is given that a:b=1:\sqrt3

=>\frac{sin A}{sinB}=\frac{1}{\sqrt3}

=>\frac{sin A}{sin60^{\circ}}=\frac{1}{\sqrt3}

=>sin A=\frac{1}{2}

=>\angle A=30^{\circ},\angle C=90^{\circ}

sin30^{\circ}=\frac{BC}{AB}

=>AB \cdot sin30^{\circ}=BC

=>\frac{1}{2}\times4=BC

=>2=BC

cos30^{\circ}=\frac{AC}{AB}

=>AB \cdot cos30^{\circ}=AC

=>\frac{\sqrt3}{2}\times4=AC

=>2\sqrt3=AC

Now, Area of  \Delta ABC=\frac{1}{2}\times AC\times BC 

                                       =\frac{1}{2}\times 2\sqrt3\times 2

                                      = 2\sqrt3 \: \: cm^{2}

 

 

 


Option 1)

\frac{2}{\sqrt3}

Option 2)

4\sqrt3

Option 3)

2\sqrt3

Option 4)

\frac{4}{\sqrt3}

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