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Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

If AB > AC ...

Question

If $A B > A C > B C,$ arrange the angles $x, y$ and $z$ in decending order of their values.

y < z < x

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x > y > z

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z > y > x

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Incomplete question

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Solution

The correct option is B $x > y > z$
Given $A B > A C > B C,$
This means that angles opposite these sides are in the same order.
So $,$ $∠ C > ∠ B > ∠ A$
$∠ z = 180 - ∠ C ∠ y = 180 - ∠ B ∠ x = 180 - ∠ A$
From the above statements it is clear that $x > y > z .$
Hence $,$ $B$ is the answer.

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In the same figure, if $y > x > z;$ arrange sides $A B, B C$ and $A C$ in descending order according their lengths.

k

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z = x^{3} sin (\frac{x}{y})

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = k z

k =

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Z = \frac{x^{3} + y^{3}}{x - y}

\frac{x \frac{\partial^{2} z}{\partial x^{2}} + y \frac{\partial^{2} z}{\partial x \partial y}}{\frac{\partial z}{\partial x}} = k

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x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = k

D

z = {sin}^{- 1} (\frac{\sqrt{x} + \sqrt{y}}{x - y})

x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = k tan z

k =

a = \frac{x}{(y - z)}, b = \frac{y}{(z - x)}, c = \frac{z}{(x - y)}

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a b + b c + c a

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