Question

In the figure given A is the centre of the circle with $\angle $BAC = \[85{}^\circ \]and $\angle $BAD=\[115{}^\circ \], then find $\angle $CBD.

Answer 1

Hint:We know what are the radii here, that is all line segments with one point on center and one point on the circle is a line segment which is AC, AD, AB is radii. We know that all radii are of equal length. Then the triangles CAB and BAD are isosceles triangles because of AC = AB and AD = AB. Now you can find the angles as those are isosceles triangles.

Complete step-by-step solution:
Given that $\angle $BAC = \[85{}^\circ \] and $\angle $BAD=\[115{}^\circ \]
We have known that the triangle CAB is isosceles which implies
$\angle $BCA = $\angle $ABC = x ;
We know that as CAB is a triangle $\angle $BCA +$\angle $CBA + $\angle $BAC = \[180{}^\circ \]
$\Rightarrow 2x + \angle $ BAC = \[180{}^\circ \]
$\Rightarrow 2x + 85{}^\circ $= \[180{}^\circ \] [As it is given that $\angle $BAC =\[85{}^\circ \]]
$x = {{\left( \dfrac{95}{2} \right)}^{{}^\circ }} = \angle CBA$
We have known that the triangle ABD is isosceles which implies
$\angle $BDA = $\angle $ABD = y ;
We know that as CAB is a triangle $\angle $BAD +$\angle $ADB + $\angle $DBA = \[180{}^\circ \]
$ 2y + \angle BAD$ = \[180{}^\circ \]
$\Rightarrow 2y + 115{}^\circ = 180{}^\circ $ [As it is given that $\angle BAD$ =\[115{}^\circ \]]
$ \Rightarrow y = {{\left( \dfrac{65}{2} \right)}^{{}^\circ }} = \angle ABD$
The question asked is to find $\angle $CBD which is equal to $\angle $CBA + $\angle $ ABD
Which implies
$\angle CBD = \angle CBA +\angle ABD$
$\Rightarrow \angle CBD = x + y$
$\Rightarrow\angle CBD$ = \[{{\left( \dfrac{65}{2} \right)}^{{}^\circ }}\]+ \[{{\left( \dfrac{95}{2} \right)}^{{}^\circ }}\]= \[80{}^\circ \]

Note: We can also do this in another model which is , the angle between two point making lines with the centre is twice the angle between the same two point making the lines with any point on upper sector of the circle , which implies $\angle $CAD(outer one) = 2($\angle $CBD). So as we know that angle in a point in \[360{}^\circ \] ,
 which implies
$\angle CAD$(outer one) + $\angle $CAB + $\angle $DAB = \[360{}^\circ \]
$\Rightarrow \angle CAD$(outer one) + \[85{}^\circ \] + \[115{}^\circ \] = \[360{}^\circ \]
$\Rightarrow \angle CAD$(outer one) = \[160{}^\circ \]
Which implies $\angle $CBD = $\dfrac{\angle CAD\text{(outer one)}}{2}$ = \[{{\left( \dfrac{160}{2} \right)}^{{}^\circ }}\text{ }=\text{ }80{}^\circ \]