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Circles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSCircles1 Mark
Question
Write True or False and justify your answer in the following:
In Fig. if AOB is a diameter and $\angle\text{ADC}=120^\circ,$ then $\angle\text{CAB}=30^\circ.$
✓

Answer

True.
Solution:
Join CA and CB.

Since, ADCB is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{CBA}=180^\circ$ [sum of opposite angles of cyclic quadrilateral is 180°]
$\Rightarrow\angle\text{CBA}=180^\circ-120^\circ=60^\circ\ \ [\therefore\angle\text{ADC}=120^\circ]$
In $\triangle\text{ACB,}\ \angle\text{CAB} + \angle\text{CBA} + \angle\text{ACB} = 180^\circ$ [by angle sum property of a triangle]
$\angle\text{CAB}+60^\circ+90^\circ=180^\circ$ $\big[$triangle formed from diameter to the circle is 90° i.e., $\angle\text{ACB}=90^\circ\big]$
$\Rightarrow\angle\text{CAB}=180^\circ-150^\circ=30^\circ.$

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