Question

If the angles of a triangle $ABC$ are in $A.P$.,
$a=2,c=4$, then $b=$

• A. $2\sqrt{3}$
• B. $\sqrt{21}$
• C. $8$
• D. $14$

Answer 1

The correct option is A $2\sqrt{3}$

Given $\to$

Angles are in AP. and a = 2, c = 4

to find $\to$ b = ?

By Sine Rule $\to$

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{C}{sinC}$

$\frac{2}{sin\theta }=\frac{x}{sin2\theta }=\frac{4}{sin3\theta }$

$sin2A=2sinAcosA$

$\frac{2}{sin\theta }=\frac{x}{2sin\theta cos\theta }$

$cos\theta =\frac{x}{4}$

Now, Cosine Rule $\to$

$cosA=\frac{b^2+c^2-a^2}{2bc}$

$cos\theta =\frac{16+x^2-4}{8x}$

$\frac{x}{4}=\frac{x^2+12}{8x}$

$2x^2=x^2+12$

$x^2=12$

$x=\pm 2\sqrt{3}$

As side can't be negative

so, $b=2\sqrt{3}$