Question

The diagonals of a parallelogram are inclined to each other at an angle of ${45}^0$, while its sides a and b (a > b) are inclined to each other at an angle of ${30}^0$, then the value of $\frac{a}{b}$ is

• A. $2\cos {36}^0$
  
• B. $\sqrt{\frac{3+\sqrt{5}}{4}}$
• C. $\frac{3+\sqrt{5}}{4}$
• D. $\frac{\sqrt{5}+1}{2}$

Answer 1

Answer: (D)

$\frac{\sqrt{5}+1}{2}$

Area of ABCD$=ab\sin 30$

Area of ABCD$=\frac{1}{2}{d}_1{d}_2\sin 45$

$⟹ab\sin 30=\frac{1}{2}{d}_1{d}_2\sin 45$

$\therefore d_1{d}_2=\sqrt{2}ab⟹{{(d}_1{d}_2)}^2=2a^2{b}^2$

By cosine rule,

$d_1^2=a^2+b^2-2ab\cos 30=a^2+b^2-\sqrt{3}ab{d}_2^2=a^2+b^2-2ab\cos 150=a^2+b^2+\sqrt{3}ab⟹{{(d}_1{d}_2)}^2=(a^2+b^2-\sqrt{3}ab)(a^2+b^2+\sqrt{3}ab)=2a^2{b}^2\therefore {(a^2+b^2)}^2-3a^2{b}^2=2a^2{b}^2{⟹a}^4+b^4+2a^2{b}^2-3a^2{b}^2=2a^2{b}^2{⟹a}^4+b^4-3a^2{b}^2=0⟹\frac{a^4}{b^4}+1-\frac{3a^2}{b^2}=0⟹{\left(\frac{a^2}{b^2}\right)}^2-\left(\frac{3a^2}{b^2}\right)+1=0$

$($quadratic in $\frac{a^2}{b^2})\therefore \frac{a^2}{b^2}=\frac{3\pm \sqrt{5}}{2}⟹\frac{a^2}{b^2}=\frac{3+\sqrt{5}}{2}(\because a>b)\therefore \frac{a^2}{b^2}=\frac{6+2\sqrt{5}}{4}=\frac{({1+\sqrt{5})}^2}{4}⟹\frac{a}{b}=\frac{1+\sqrt{5}}{2}$