Find the angle $ \alpha $ .
Answer
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Hint: We will use the property of the exterior angle to obtain an equation with the angles $ \alpha $ and $ \beta $ . Then, we will look at the right-angled triangle $ \Delta \text{DEB} $ . We will use the definition of the tangent function in this triangle to find the value of $ \beta $ . Then we will substitute this value in the previously obtained equation to find the value of the angle $ \alpha $ .
Complete step by step answer:
From the figure, we are given that $ \angle \text{CDB}=65{}^\circ $ and $ \angle \text{DEB}=90{}^\circ $ . We are also given that $ \text{ED}=21 $ and $ \text{EB}=50 $ . We can see in the figure that $ \angle \text{CDB} $ is an exterior angle of $ \Delta \text{DAB} $ . We know that the exterior angle is equal to the sum of its interior angle. Hence, we get the following equation,
$ \angle \text{DAB}+\angle \text{DBA}=\angle \text{CDB} $
Substituting the values of these three angles from the figure, we get
$ \alpha +\beta =65{}^\circ ....(i) $
Now, let us consider the right angled triangle $ \Delta \text{DEB} $ . We know the definition of the tangent function as $ \tan \theta =\dfrac{\text{Opposite}}{\text{Adjacent}} $ . Using this definition, we will find the value of the tangent function for the angle $ \beta $ in the following manner,
$ \tan \beta =\dfrac{\text{ED}}{\text{EB}} $
Substituting the values $ \text{ED}=21 $ and $ \text{EB}=50 $ in the above equation, we get
\[\begin{align}
& \tan \beta =\dfrac{21}{50} \\
& \therefore \tan \beta =0.42 \\
\end{align}\]
Hence, we get the value of angle $ \beta $ as $ \beta ={{\tan }^{-1}}\left( 0.42 \right) $ . Therefore, we have $ \beta =22.78{}^\circ $ . Now, substituting this value in equation $ (i) $ , we get
$ \begin{align}
& \alpha +22.78{}^\circ =65{}^\circ \\
& \Rightarrow \alpha =65{}^\circ -22.78{}^\circ \\
& \therefore \alpha =42.22{}^\circ \\
\end{align} $
Note:
It is important that we understand the geometry of a given figure. The concepts of exterior angles and interior angles are useful for such type of questions. We should know the definition of trigonometric functions since they can be used to obtain the values of angles by looking at the inverse trigonometric functions.
Complete step by step answer:
From the figure, we are given that $ \angle \text{CDB}=65{}^\circ $ and $ \angle \text{DEB}=90{}^\circ $ . We are also given that $ \text{ED}=21 $ and $ \text{EB}=50 $ . We can see in the figure that $ \angle \text{CDB} $ is an exterior angle of $ \Delta \text{DAB} $ . We know that the exterior angle is equal to the sum of its interior angle. Hence, we get the following equation,
$ \angle \text{DAB}+\angle \text{DBA}=\angle \text{CDB} $
Substituting the values of these three angles from the figure, we get
$ \alpha +\beta =65{}^\circ ....(i) $
Now, let us consider the right angled triangle $ \Delta \text{DEB} $ . We know the definition of the tangent function as $ \tan \theta =\dfrac{\text{Opposite}}{\text{Adjacent}} $ . Using this definition, we will find the value of the tangent function for the angle $ \beta $ in the following manner,
$ \tan \beta =\dfrac{\text{ED}}{\text{EB}} $
Substituting the values $ \text{ED}=21 $ and $ \text{EB}=50 $ in the above equation, we get
\[\begin{align}
& \tan \beta =\dfrac{21}{50} \\
& \therefore \tan \beta =0.42 \\
\end{align}\]
Hence, we get the value of angle $ \beta $ as $ \beta ={{\tan }^{-1}}\left( 0.42 \right) $ . Therefore, we have $ \beta =22.78{}^\circ $ . Now, substituting this value in equation $ (i) $ , we get
$ \begin{align}
& \alpha +22.78{}^\circ =65{}^\circ \\
& \Rightarrow \alpha =65{}^\circ -22.78{}^\circ \\
& \therefore \alpha =42.22{}^\circ \\
\end{align} $
Note:
It is important that we understand the geometry of a given figure. The concepts of exterior angles and interior angles are useful for such type of questions. We should know the definition of trigonometric functions since they can be used to obtain the values of angles by looking at the inverse trigonometric functions.
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