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The below figure is given in the above question.

We are asked to find angle $ \alpha $ in the above figure.

First of all, we are going to find angle $ \beta $ . As we can see that $ \Delta DEC $ is a right triangle and using the trigonometric ratios we can find angle $ \beta $ by using $ \tan \beta $ .

We know that,

$ \tan \beta =\dfrac{\text{Perpendicular}}{\text{Base}} $

In the above figure, perpendicular is 21 and base is 50 so substituting these values in the above equation we get,

$ \tan \beta =\dfrac{21}{50} $

Taking $ {{\tan }^{-1}} $ on both the sides we get,

$ \beta ={{\tan }^{-1}}\left( \dfrac{21}{50} \right) $

From the given figure, we can see that ADB is a linear pair so the sum of angles lying on this line is $ {{180}^{\circ }} $ .

Adding all the angles lie on the straight line ADB we get,

$ \begin{align}

& \angle ADC+\angle CDE+\angle EDB={{180}^{\circ }} \\

& \Rightarrow {{65}^{\circ }}+\angle CDE+\angle EDB={{180}^{\circ }} \\

\end{align} $

Subtracting $ {{65}^{\circ }} $ on both the sides we get,

$ \begin{align}

& \angle CDE+\angle EDB={{180}^{\circ }}-{{65}^{\circ }} \\

& \Rightarrow \angle CDE+\angle EDB={{115}^{\circ }} \\

\end{align} $

Now, $ \angle CDE={{90}^{\circ }}-\beta $ so substituting this angle in the above equation we get,

$ \begin{align}

& {{90}^{\circ }}-\beta +\angle EDB={{115}^{\circ }} \\

& \Rightarrow \angle EDB={{115}^{\circ }}-{{90}^{\circ }}+\beta \\

& \Rightarrow \angle EDB={{25}^{\circ }}+\beta \\

\end{align} $

Now, $ \Delta DEB $ is a right triangle so angle $ \alpha $ is equal to $ {{90}^{\circ }}-\angle EDB $ we get,

$ \alpha ={{90}^{\circ }}-\angle EDB $

Substituting the value of angle EDB from the above equations we get,

$ \begin{align}

& \alpha ={{90}^{\circ }}-\left( {{25}^{\circ }}+\beta \right) \\

& \Rightarrow \alpha ={{65}^{\circ }}-\beta \\

\end{align} $

Substituting the value of angle $ \beta ={{\tan }^{-1}}\left( \dfrac{21}{50} \right) $ we get,

$ \alpha ={{65}^{\circ }}-{{\tan }^{-1}}\left( \dfrac{21}{50} \right) $

Hence, we have got the value of $ \alpha $ as $ {{65}^{\circ }}-{{\tan }^{-1}}\left( \dfrac{21}{50} \right) $ .