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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)

best_answer

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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