Question

In $\Delta ABC,$ the ratio $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ is always equal to:
(where for $△ABC$ usual notations are used and $h_1,h_2,h_3$ are altitudes from respective vertices.)

• A. $\frac{(abc{)}^{\frac{1}{2}}}{(h_1{h}_2{h}_3{)}^{\frac{1}{6}}}$
• B. $\frac{(abc{)}^{\frac{4}{3}}}{\left(h_1{h}_2{h}_3\right)}$
• C. $\frac{(abc{)}^{\frac{1}{3}}}{(h_1{h}_2{h}_3{)}^{\frac{2}{3}}}$
• D. $\frac{(abc{)}^{\frac{2}{3}}}{(h_1{h}_2{h}_3{)}^{\frac{1}{3}}}$

Answer 1

The correct option is D $\frac{(abc{)}^{\frac{2}{3}}}{(h_1{h}_2{h}_3{)}^{\frac{1}{3}}}$

We know, in a $\Delta ABC,$ using sine law, the ratio $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$
Also $abc=4R\Delta \Rightarrow 2R=\frac{abc}{2\Delta }$
With $h_1,h_2,h_3$ being altitudes from the vertices $A,B,C$ respectively, we have: $\frac{1}{2}a{h}_1=\frac{1}{2}b{h}_2=\frac{1}{2}c{h}_3=\Delta \Rightarrow h_1=\frac{2\Delta }{a},h_2=\frac{2\Delta }{b},h_3=\frac{2\Delta }{c}\therefore h_1{h}_2{h}_3=\frac{8{\Delta }^3}{abc}$
So we have
$\frac{(abc{)}^{\frac{2}{3}}}{(h_1{h}_2{h}_3{)}^{\frac{1}{3}}}=\frac{(abc{)}^{\frac{2}{3}}}{{\left(\frac{8{\Delta }^3}{abc}\right)}^{\frac{1}{3}}}=\frac{\left(abc{)}^{\frac{2}{3}}\right(abc{)}^{\frac{1}{3}}}{2\Delta }\Rightarrow \therefore 2R=\frac{\left(abc\right)}{2\Delta }=\frac{(abc{)}^{\frac{2}{3}}}{(h_1{h}_2{h}_3{)}^{\frac{1}{3}}}$