In Delta ABC,angle B is a right angle. If AC | vedclass

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(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.We are given the hypotenuse $AC = 20$ and the adjacent side to $\angle ACB$,which is $BC = 10$.Using the t

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$60$

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(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.We are given the hypotenuse $AC = 20$ and the adjacent side to $\angle ACB$,which is $BC = 10$.Using the trigonometric ratio for cosine:$\cos(\angle ACB) = \frac{	ext{Adjacent}}{	ext{Hypotenuse}} = \frac{BC}{AC}$$\cos(\angle ACB) = \frac{10}{20} = \frac{1}{2}$Since $\cos(60^{\circ}) = \frac{1}{2}$,we have $m\angle ACB = 60^{\circ}$.

In $\Delta ABC$,$\angle B$ is a right angle. If $AC = 20$ and $BC = 10$,then $m\angle ACB = \ldots$ (in $^{\circ}$)

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