Question

If in a triangle abc angle a=pi/6 and b:c=2:sqrt3, find angle b

Answer 1

$\textbf{Given:}$

$\text{In \triangle\,ABC , A=\frac{\pi}{6} and b:c=2:\sqrt{3} }$

$\textbf{To find:}$

$\text{Angle B}$

$\textbf{Solution:}$

$\text{Consider,}$

$b:c=2:\sqrt{3}$

$\implies\,b=2k\,\text{and}\,c=\sqrt{3}k$

$\text{By cosine formula}$

$a^2=b^2+c^2-2bc\,cosA$

$a^2=(2k)^2+(\sqrt{3}k)^2-2(2k)(\sqrt{3}k)\,cos\frac{\pi}{6}$

$a^2=4k^2+3k^2-2(2k)(\sqrt{3}k)(\frac{\sqrt{3}}{2})$

$a^2=7k^2-6k^2$

$a^2=k^2$

$\implies\,a=k$

$\text{By Sine formula,}$

$\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}$

$\implies\dfrac{a}{sinA}=\dfrac{b}{sinB}$

$\implies\dfrac{k}{sin\frac{\pi}{6}}=\dfrac{2k}{sinB}$

$\implies\dfrac{k}{\frac{1}{2}}=\dfrac{2k}{sinB}$

$\implies\dfrac{2k}{1}=\dfrac{2k}{sinB}$

$\implies\,sinB=1$

$\implies\bf\,B=\dfrac{\pi}{2}$

$\textbf{Answer:}$

$\textbf{Angle B is \bf\dfrac{\pi}{2} }$

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Answer 2

Answer:

Step-by-step explanation: