Question
If the sides of a triangle are in G.P. and the largest angle is twice the smallest angle then the common ratio, which is greater than 1, lies in the interval
A.
$$\left( {1,\sqrt 3 } \right)$$
B.
$$\left( {1,\root 4 \of 3 } \right)$$
C.
$$\left( {1,\frac{{\sqrt 5 + 1}}{2}} \right)$$
D.
None of these
Answer :
$$\left( {1,\root 4 \of 3 } \right)$$
Solution :
$$b = ar,c = a{r^2},$$ where $$r > 1.$$
From the question, $$C = 2A.\,\,{\text{So}},B = \pi - A - C = \pi - 3A.$$
$$\eqalign{
& \therefore \,\,\frac{a}{{\sin \,A}} = \frac{b}{{\sin \,B}} = \frac{c}{{\sin \,C}} \cr
& \Rightarrow \,\,\frac{1}{{\sin \,A}} = \frac{r}{{\sin \,3A}} = \frac{{{r^2}}}{{\sin \,2A}} \cr
& \therefore \,\,{r^2} = 2\cos \,A\,\,{\text{and}}\,\,r = \frac{{\sin \,3A}}{{\sin \,A}} = 3 - 4{\sin ^2}A = 4{\cos ^2}A - 1 \cr
& \therefore \,\,r = {r^4} - 1\,\,\,{\text{or,}}\,\,{r^4} = 1 + r. \cr} $$
Among $$\sqrt 3 ,\root 4 \of 3 ,\frac{{\sqrt 5 + 1}}{2}$$ we find $$\root 4 \of 3 $$ is the smallest because $${\left( {\sqrt 3 } \right)^4} = 9,{\left( {\root 4 \of 3 } \right)^4} = 3,{\left( {\frac{{\sqrt 5 + 1}}{2}} \right)^4} = {\left( {\frac{{3 + \sqrt 5 }}{2}} \right)^2} = \frac{{14 + 6\sqrt 5 }}{4} > 3.$$
Now, putting $$r = \root 4 \of 3 ,$$ we have $$3 > 1 + \root 4 \of 3 ,$$ implying $${\root 4 \of 3 }$$ does not satisfy $${r^4} = 1 + r.\,\,{\text{So, }}r < \root 4 \of 3 .$$