In ∆ABC, AD and BE are the medians drawn through the angular points A and B respectively. ∠DAB=2∠ABE=36∘ and AD=6 units then circum radius of the triangle equal to

By Applying sine rile in ∆AGB4sin⁡18°=csin⁡126°(∵G divides AD in the ratio 2:1 then AG=4,GD=2)⇒4sin⁡18°=ccos⁡36°⇒c=4cos⁡36°sin⁡18°=4(5+145-14)(∵sin⁡18°=14(5−1),cos⁡36°=14(5+1))⇒c=4(5+15-1)=4((5+1)2(5-1)(5 +1))=6+25Now from sine rulecsin⁡C=2R⇒R=c2sin⁡C⇒R=(6+252)cosec⁡C ⇒R=(5+3)cosec⁡C

In ∆ABC, AD and BE are the medians drawn through the angular points A and B respectively. ∠DAB=2∠ABE=36∘ and AD=6 units then circum radius of the triangle equal to

By Applying sine rile in ∆AGB4sin⁡18°=csin⁡126°(∵G divides AD in the ratio 2:1 then AG=4,GD=2)⇒4sin⁡18°=ccos⁡36°⇒c=4cos⁡36°sin⁡18°=4(5+145-14)(∵sin⁡18°=14(5−1),cos⁡36°=14(5+1))⇒c=4(5+15-1)=4((5+1)2(5-1)(5 +1))=6+25Now from sine rulecsin⁡C=2R⇒R=c2sin⁡C⇒R=(6+252)cosec⁡C ⇒R=(5+3)cosec⁡C

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In $∆$ ABC, AD and BE are the medians drawn through the angular points A and B respectively. $∠ DAB = 2 ∠ ABE = 36^{\circ} and AD = 6 units$ then circum radius of the triangle equal to

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$(3 - \sqrt{5}) cosec C^{1}$

$(3 + \sqrt{5}) cosec C$

$2 (3 - \sqrt{5}) cosec C^{1}$

$2 (3 + \sqrt{5}) cosec C^{1}$

answer is B.

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Detailed Solution

By Applying sine rile in $∆$ AGB

$\begin{array}{l} \frac{4}{\sin 18 °} = \frac{c}{\sin 126 °} (∵ G d i v i d e s A D i n t h e r a t i o 2 : 1 t h e n A G = 4, G D = 2) \\ \Rightarrow \frac{4}{\sin 18 °} = \frac{c}{\cos 36 °} \\ \Rightarrow c = 4 (\frac{\cos 36 °}{\sin 18 °}) = 4 (\frac{\frac{\sqrt{5} + 1}{4}}{\frac{\sqrt{5} - 1}{4}}) (∵ \sin 18 ° = \frac{1}{4} (\sqrt{5} - 1), \cos 36 ° = \frac{1}{4} (\sqrt{5} + 1)) \\ \Rightarrow c = 4 (\frac{\sqrt{5} + 1}{\sqrt{5} - 1}) = 4 (\frac{{(\sqrt{5} + 1)}^{2}}{(\sqrt{5} - 1) (\sqrt{5} + 1)}) = 6 + 2 \sqrt{5} \\ N o w f r o m \sin e r u l e \frac{c}{\sin C} = 2 R \\ \Rightarrow R = \frac{c}{2 \sin C} \\ \Rightarrow R = (\frac{6 + 2 \sqrt{5}}{2}) cosec C \\ \Rightarrow R = (\sqrt{5} + 3) cosec C \end{array}$