In a Triangle ABC, if Cot A : Cot B : Cot C = 1:4:15 Then The Greates

1.	In a Triangle ABC, if Cot A : Cot B : Cot C = 1:4:15 Then The Greatest Angle is?
Answer» cot A : cot B : cot C = 1 : 4 : 15 Let cot A = x, cot B = 4x and cot C = 15x ⇒ A = cot^-1x, B = cot^-14x + cot^-115x = 180° (\(\because\) cot^-1 x + cot^-1 y = cot^-1(\(\frac{xy-1}{x+y}\))) ⇒ cot^-1\(\left(\cfrac{\frac{4x^2-1}{5x}\times15-1}{\frac{4x^2-1}{5x}+15x}\right)\)= 180° ⇒ cot^-1\(\left(\cfrac{\frac{15x(4x^2-1)-5x}{5x}}{\frac{4x^2-1+75x^2}{5x}}\right)\)= 180° ⇒ \(\frac{15x(4x^2-1)-5x}{79x^2-1}\) = cot 180° = \(\frac{cos180^{\circ}}{sin180^{\circ}}\) = \(\frac{-1}0\) \(\therefore\) 79x² - 1 = 0 ⇒ x = \(\frac{1}{\sqrt{79}}\) Now, A = cot^-1x = cot^-1\((\frac1{\sqrt{79}})\) B = cot^-14x = cot^-1\((\frac{4}{\sqrt{79}})\) C = cot^-115x = cot^-1\((\frac{15}{\sqrt{79}})\) A > B > C (\(\because\) cot^-1 x is a decreasing function and x < 4x < 15x) Greatest angle is A.

In a Triangle ABC, if Cot A : Cot B : Cot C = 1:4:15 Then The Greatest Angle is?

Answer»

cot A : cot B : cot C = 1 : 4 : 15

Let cot A = x, cot B = 4x and cot C = 15x

⇒ A = cot^-1x, B = cot^-14x + cot^-115x = 180°

(\(\because\) cot^-1 x + cot^-1 y = cot^-1(\(\frac{xy-1}{x+y}\)))

⇒ cot^-1\(\left(\cfrac{\frac{4x^2-1}{5x}\times15-1}{\frac{4x^2-1}{5x}+15x}\right)\)= 180°

⇒ cot^-1\(\left(\cfrac{\frac{15x(4x^2-1)-5x}{5x}}{\frac{4x^2-1+75x^2}{5x}}\right)\)= 180°

⇒ \(\frac{15x(4x^2-1)-5x}{79x^2-1}\) = cot 180° = \(\frac{cos180^{\circ}}{sin180^{\circ}}\) = \(\frac{-1}0\)

\(\therefore\) 79x² - 1 = 0

⇒ x = \(\frac{1}{\sqrt{79}}\)

Now, A = cot^-1x = cot^-1\((\frac1{\sqrt{79}})\)

B = cot^-14x = cot^-1\((\frac{4}{\sqrt{79}})\)

C = cot^-115x = cot^-1\((\frac{15}{\sqrt{79}})\)

A > B > C

(\(\because\) cot^-1 x is a decreasing function and x < 4x < 15x)

Greatest angle is A.

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