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Question: In the given figure, \(\angle AOB=90{}^\circ \) and \(\angle ABC=30{}^\circ \), Then \(\angle CAD=?\...

In the given figure, AOB=90\angle AOB=90{}^\circ and ABC=30\angle ABC=30{}^\circ , Then CAD=?\angle CAD=?

(a) 3030{}^\circ
(b) 4545{}^\circ
(c) 6060{}^\circ
(d) 9090{}^\circ

Explanation

Solution

In triangle OAB, OAB=OBA\angle OAB=\angle OBA, calculate OBA\angle OBA. Assume point “M” as shown in the diagram. MBA=30\angle MBA=30{}^\circ (given). Find MBC\angle MBC by subtracting MBA\angle MBA from OBA\angle OBA. As OC=OB=r. So, OCM=OBM\angle OCM=\angle OBM. Now calculate COB\angle COBand then calculate CAO\angle CAO by using the fact that “angle subtended by a chord at the center of a circle is double of the angle subtended by the chord at any point lying on the circle.”

Complete step-by-step solution:
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All the points are specified. Assume point of intersection of OA and BC on “M”.
In OAB,OA=OB\vartriangle OAB,\,OA=OB=radius of the circle.
We know that if two sides is a triangle are equal, their corresponding angles will also be equal.
i.e. OAB=OBA\angle OAB=\angle OBA
let OAB=OBA=x\angle OAB=\angle OBA=x
By angle sum property of a circle, sum of the angles of a \vartriangle is 180180{}^\circ .
In \vartriangle OAB-
OAB+OBA+AOB=180 x+x+90=180 2x=18090 2x=90 x=45 \begin{aligned} & \angle OAB+\angle OBA+\angle AOB=180{}^\circ \\\ & \Rightarrow x+x+90{}^\circ =180{}^\circ \\\ & \Rightarrow 2x=180{}^\circ -90{}^\circ \\\ & \Rightarrow 2x=90{}^\circ \\\ & \Rightarrow x=45{}^\circ \\\ \end{aligned}
So, OBA=OAB=45\angle OBA=\angle OAB=45{}^\circ
From diagram,
OBA=OBM+MBA 45=OBM+30 OBM=4530 OBM=15 \begin{aligned} & \angle OBA=\angle OBM+\angle MBA \\\ & \Rightarrow 45{}^\circ =\angle OBM+30{}^\circ \\\ & \Rightarrow \angle OBM=45{}^\circ -30{}^\circ \\\ & \Rightarrow \angle OBM=15{}^\circ \\\ \end{aligned}
In OCB\vartriangle OCB
OC=OB (both are radius of the circle)
So, corresponding angles will also be equal in the \vartriangle .
So, OCB=OBC\angle OCB=\angle OBC
We have calculated above that OBM=15\angle OBM=15{}^\circ
OBC=15\Rightarrow \angle OBC=15{}^\circ (see diagram)
So, OBC=OBM=15\Rightarrow \angle OBC=\angle OBM=15{}^\circ
Now, in \vartriangle OBM-
MOB=90(given) OBM=15(calculatedabove) \begin{aligned} & \angle MOB=90{}^\circ \,(given) \\\ & \angle OBM=15{}^\circ \,(calculated\,above) \\\ \end{aligned}
By angle sum property of triangle-
MOB+OBM+OMB=180 90+15+OMB=180 OMB=180(90+15) OMB=75 \begin{aligned} & \angle MOB+\angle OBM+\angle OMB=180{}^\circ \\\ & 90{}^\circ +15{}^\circ +\angle OMB=180{}^\circ \\\ & \angle OMB=180{}^\circ -(90{}^\circ +15{}^\circ ) \\\ & \angle OMB=75{}^\circ \\\ \end{aligned}
By linear property straight lines on line BC-
CMO+OMB=180 CMO+75=180 [OMB=75,calculatedabove] CMO=18075 CMO=105 \begin{aligned} & \angle CMO+\angle OMB=180{}^\circ \\\ & \angle CMO+75{}^\circ =180{}^\circ \\\ & \left[ \angle OMB=75{}^\circ ,\,calculated\,above \right] \\\ & \angle CMO=180{}^\circ -75{}^\circ \\\ & \angle CMO=105{}^\circ \\\ \end{aligned}
Now in OCM\vartriangle OCM-
OCM=15(calculatedabove) OMC=105(calculatedabove) \begin{aligned} & \angle OCM=15{}^\circ (calculated\,above) \\\ & \angle OMC=105{}^\circ (calculated\,above) \\\ \end{aligned}
By angle sum property of a \vartriangle
OCM+OMC+COM=180 15+105+COM=180 120+COM=180 COM=180120 COM=60 \begin{aligned} & \angle OCM+\angle OMC+\angle COM=180{}^\circ \\\ & \Rightarrow 15{}^\circ +105{}^\circ +\angle COM=180{}^\circ \\\ & \Rightarrow 120{}^\circ +\angle COM=180{}^\circ \\\ & \Rightarrow \angle COM=180{}^\circ -120{}^\circ \\\ & \Rightarrow \angle COM=60{}^\circ \\\ \end{aligned}
From the diagram, we can see that
COB=COM+MOB COM=60(calcullatedabove) MOB=90(given) So,COB=60+90 COB=150 \begin{aligned} & \angle COB=\angle COM+\angle MOB \\\ & \angle COM=60{}^\circ \,(calcullated\,above) \\\ & \angle MOB=90{}^\circ \,(given) \\\ & So,\angle COB=60{}^\circ +90{}^\circ \\\ & \Rightarrow \angle COB=150{}^\circ \\\ \end{aligned}
We know that the angle subtended by a chord of a circle at its center is double the angle subtended by the chord at any point on its circumference.
We can see the diagram that chord BC subtends COB\angle COB at the center of the circle and angle CAB\angle CAB at the circumference.
So,
COB=2×CAB 150=2×CAB[COB=150ascalculatedabove] CAB=1502=75 \begin{aligned} & \angle COB=2\times \angle CAB \\\ & \Rightarrow 150{}^\circ =2\times \angle CAB\,\,\,\left[ \angle COB=150{}^\circ \,as\,calculated\,above \right] \\\ & \Rightarrow \angle CAB=\dfrac{150{}^\circ }{2}=75{}^\circ \\\ \end{aligned}
From the diagram, we can see that CAB=CAD+OAB\angle CAB=\angle CAD+\angle OAB
CAB=75(calculatedabove) OAB=45(calculatedabove) \begin{aligned} & \angle CAB=75{}^\circ \,(calculated\,above) \\\ & \angle OAB=45{}^\circ \,(calculated\,above) \\\ \end{aligned}
So,
75=CAO+45 CAD=7545=30 \begin{aligned} & 75{}^\circ =\angle CAO+45{}^\circ \\\ & \Rightarrow \angle CAD=75{}^\circ -45{}^\circ =30{}^\circ \\\ \end{aligned}
Hence the required value of CAD\angle CAD will be 3030{}^\circ and option (a) is the correct answer.

Note: We have used a theorem in the solution that the angle subtended by a chord at the center is double of the angle subtended by it on its circumference.

Given chord PQ of a circle subtending POQ at the centre O and PAQ at a point A n the circumference of the circle.
To prove:
POQ=2PAQ\angle POQ=2\angle PAQ
Join AM such that AM is bisecting PAQ\angle PAQ
Let, PAD=DAQ=θ\angle PAD=\angle DAQ=\theta
In ADQOA=DQ(Bothequaltoradius)\vartriangle ADQ-OA=DQ\,(Both\,equal\,to\,radius)
So, corresponding angles will also be equal.
i.e. OAQ=OQA\angle OAQ=\angle OQA
So, OQA=θ\angle OQA=\theta
Now by exterior angle sum property- (in AOQ\vartriangle AOQ-
OAQ+DQA=MOQ θ+θ=MOq MOQ=2θ \begin{aligned} & \angle OAQ+\angle DQA=\angle MOQ \\\ & \Rightarrow \theta +\theta =\angle MOq \\\ & \Rightarrow \angle MOQ=2\theta \\\ \end{aligned}
Similarly, in APO\vartriangle APO- (exterior angles property)
POM=2θ Noq,POQ=2θ+2θ=4θ(Seediagram) andPAQ=θ+θ=2 \begin{aligned} & \angle POM=2\theta \\\ & Noq,\angle POQ=2\theta +2\theta =4\theta \,(See\,diagram) \\\ & and\,\angle PAQ=\theta +\theta =2 \\\ \end{aligned}
Hence, POQ=2PAQ\angle POQ=2\angle PAQ