Question

In $\Delta ABC,ifA:B:C=1:5:6then$ find the value of $\sin A:\sin B:\sin C$

• A. $\left(\sqrt{3}-1\right):2\sqrt{2}:\left(\sqrt{3}+1\right)$
• B. $2\sqrt{2}:\left(\sqrt{3}-1\right):\left(\sqrt{3}+1\right)$
• C. $\left(\sqrt{3}-1\right):\left(\sqrt{3}+1\right):2\sqrt{2}$
• D. $\left(\sqrt{3}-1\right):\sqrt{3}:\sqrt{2}$

Answer 1

The correct option is C $\left(\sqrt{3}-1\right):\left(\sqrt{3}+1\right):2\sqrt{2}$

Given that $A:B:C=1:5:6$

We know that the sum of angles in a triangle is ${180}^0$

Let us first find each angle.

Total no.of parts= $1+5+6=12$

$A$= $\frac{1}{12}$ (${180}^0$) =${15}^0$

$B$= $\frac{5}{12}$ (${180}^0$) =${75}^0$

$C$= $\frac{6}{12}$ (${180}^0$) =${90}^0$

Hence, $\sin A:\sin B:\sin C$ = $\sin {15}^o$$:\sin {75}^o$$:\sin {90}^o$

$=\frac{\sqrt{3}-1}{2\sqrt{2}}:\frac{\sqrt{3}+1}{2\sqrt{2}}:1$

$=\left(\sqrt{3}-1\right):\left(\sqrt{3}+1\right):\left(2\sqrt{2}\right)$