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If the side length of a regular pentagon is a and length of its diagonal is b; then prove that ((a squared)/(b squared)) + ((b squared)/(a squared) ) = 3

If the side length of a regular pentagon is a and length of its diagonal is b; then prove that
((a squared)/(b squared)) + ((b squared)/(a squared) ) = 3

Grade:10

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:Hello student, please find answer to your question
Length of side = a
Length of diagonal = b
Angle b/w sides = 720
Apply cosine rule
cos(72^{0}) = \frac{a^{2}+a^{2}-b^{2}}{2.a.a}
cos(72^{0}) = \frac{2a^{2}-b^{2}}{2a^{2}}
cos(72^{0}) = 1-\frac{b^{2}}{2a^{2}}
\frac{b^{2}}{a^{2}} = 2(1-cos(72^{0}))
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{1}{2(1-cos(72^{0}))}+2(1-cos(72^{0}))
cos(72^{0}) = \frac{\sqrt{5}-1}{4}
1-cos(72^{0}) = 1-\frac{\sqrt{5}-1}{4}
1-cos(72^{0}) = \frac{5-\sqrt{5}}{4}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{4}{2(5-\sqrt{5})}+2.\frac{5-\sqrt{5}}{4}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{2(5+\sqrt{5})}{(5-\sqrt{5})(5+\sqrt{5})}+\frac{5-\sqrt{5}}{2}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{2(5+\sqrt{5})}{20}+\frac{5-\sqrt{5}}{2}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{2(5+\sqrt{5})}{20}+\frac{10(5-\sqrt{5})}{20}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = \frac{60-8\sqrt{5}}{20}
\frac{a^{2}}{b^{2}}+\frac{b^{2}}{a^{2}} = 3-\frac{8\sqrt{5}}{20}

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