Circles Test - 23

O$$Y = OZ = radius = r$$

Given $$YZ = r$$

$$\implies \triangle OYZ $$ is equilateral

$$\implies \angle YOZ = 60^\circ$$

We know that angle made by a chord on any point on the circle is half the angle made by the chord at the center

$$\implies \angle YXZ = \dfrac{\angle YOZ}{2}$$

$$\implies \angle YXZ = \dfrac{60}{2} = 30^\circ$$

Given $$ABCD$$ is a cyclic quadrilateral.

Given, $$AB$$ is the common chord of two intersecting circles.

Given, $$ PQRS$$ is a cyclic trapezium with $$PQ\parallel SR, \angle QPS={ 82 }^{ o }.$$

Given, $$ABCD$$ is a cyclic quadrilateral. $$AC$$ is its diagonal. $$\angle BAD={ 25 }^{ o }, \angle BCD={ 80 }^{ o }$$.

Here, $$ABCD$$ is a cyclic quadrilateral. 

$$\therefore  \angle BCD+\angle BAD=1{ 80 }^{ o }$$ (since the sum of the opposite  angles of a cyclic quadrilateral is $${ 180 }^{ o }$$)

$$\Rightarrow \angle BAD=1{ 80 }^{ o }-\angle BCD=1{ 80 }^{ o }-{ 80 }^{ o }=1{ 00 }^{ o }$$. 

But $$ \angle BAD=\angle BAC+\angle CAD$$

$$\therefore  \angle BAC+\angle CAD=1{ 00 }^{ o } \Rightarrow \angle CAD=1{ 00 }^{ o }-\angle BAC=1{ 00 }^{ o }-25^{ o }=75^{ o }$$.

Hence, option $$C$$ is correct.

$$ \\ Given-\\ The\quad \Delta ABC,\quad whose\quad incentre\quad is\quad I,\quad has\quad been\quad inscribed\quad in\quad a\quad cicle.\\ AI\quad is\quad produced\quad to\quad meet\quad the\quad circumference\quad at\quad D.\\ BD\quad \& \quad DC\quad have\quad been\quad joined.\\ \angle ABC={ 55 }^{ o }\quad \& \quad \angle ACB={ 65 }^{ o }.\\ To\quad find\quad out-\\ (i)\quad \angle BCD=?\quad \quad (ii)\quad \angle CBD=?\quad \quad (iii)\quad \angle DCI=?\quad \quad (iv)\quad \angle BIC=?\\ Solution-\\ \angle ABC=\angle ADC={ 55 }^{ o }\quad [since\quad both\quad the\quad angles\quad have\quad been\quad subtended\quad \\ by\quad the\quad chord\quad AC\quad to\quad the\quad circumference\quad at\quad B\quad \& \quad D].\\ Also, \quad  \angle ACB=\angle ADB={ 65 }^{ o }\quad [since\quad both\quad the\quad angles\quad have\quad been\quad subtended\quad \\ by\quad the\quad chord\quad AB\quad to\quad the\quad circumference\quad at\quad C\quad \& \quad D].\\ So,\quad \angle BDC=\angle ADC+\angle ADB={ 55 }^{ o }+{ 65 }^{ o }=120^{ o }.\\ Now,\quad reflex\angle BIC=2\angle BDC\quad (since\quad reflex\angle BIC\quad \& \quad \angle BDC\quad are\quad angles\quad \\ at\quad the\quad centre\quad \& \quad angle\quad at\quad the\quad circumference\quad at\quad D\quad subtended\quad by\\ the\quad chord\quad BC).\\ \therefore \quad reflex\angle BIC=2\times 120^{ o }=240^{ o }\Longrightarrow \angle BIC=360^{ o }-240^{ o }=120^{ o }.\\ By\quad the\quad same\quad reasoning\quad \angle BAC=\quad \frac { 1 }{ 2 } \times \angle BIC=\frac { 1 }{ 2 } \times 120^{ o }=60^{ o }.\\ Again,\quad I\quad is\quad the\quad incentre\quad of\quad \Delta ABC.\\ \therefore \quad AI,\quad BI\quad \& \quad CI\quad are\quad angular\quad bisector\quad of\quad \\ \angle BAC,\quad \angle ABC\quad \& \quad \angle ACB\quad respectively.\\ \therefore \quad \angle BAI=\frac { 1 }{ 2 } \times \angle BAC=\frac { 1 }{ 2 } \times 60^{ o }=30^{ o }=\angle CAI.\\ So,\quad \angle BCD=\angle BAI=30^{ o }\quad [since\quad \angle BCD\quad \& \quad \angle BAI\quad are\quad angles\quad \\ at\quad the\quad centre\quad \& \quad angle\quad at\quad the\quad circumference\quad at\quad A\quad subtended\quad by\\ the\quad chord\quad BD].\\ By\quad the\quad same\quad reasoning\quad \angle CBD=\angle CAI=30^{ o }.\\ Again,\quad CI\quad is\quad the\quad angular\quad bisector\quad of\quad \angle ACD.\\ \therefore \quad \angle BCI=\angle ACI=\frac {1 }{ 2 } \times 65^{ o }=32.5^{ o }.\\ \therefore \quad \angle DCI=\angle BCD+\angle BCI=32.5^{ o }+30^{ o }=62.5^{ o }.\\ So\quad \angle BCD,\quad \angle CBD,\quad \angle DCI\quad \& \quad \angle BIC\quad are \quad { 30 }^{ o },\quad { 30 }^{ o },\quad { 62.5 }^{ o }\quad \& \quad { 120 }^{ o }\quad respectively.\\ Hence,\quad option\quad B \quad is \quad correct. $$

Since $$AB$$ is diameter,

$$\angle AEB = 90^\circ$$ ...[Angle formed in a semi circle].

In $$\triangle AEB$$

$$\angle ABE = 180 - \angle EAB - \angle AEB$$ ...[Angle sum property]

$$\angle ABE = 180 – 65 -90 = 25^\circ$$.

Given, $$ED \parallel AB$$.

$$\implies \angle DEB = \angle EBA = 25^\circ$$ ...[Alternate interior angles].

Since $$EDCB$$ is a cyclic quadrilateral opposite angles are supplementary

Then, $$\angle DEB + \angle DCB  = 180^\circ $$

$$\implies \angle DCB = 180 – 25 = 155^\circ $$.

Hence, option $$D$$ is correct.

Given that $$O$$ is the center of the smaller circle and $$\angle APB=75^{\circ}$$.

$$ Given-\\ Two\quad circles\quad intersect\quad at\quad A\quad \& \quad B.\\ The\quad quadrilateral\quad PCBA\quad is\quad inscribed\quad in\quad one\quad circle\quad and\\ the\quad quadrilateral\quad EDBA\quad is\quad inscribed\quad in\quad another\quad circle\\ such\quad that\quad the\quad points\quad P,\quad A\quad \& \quad B\quad are\quad collinear\quad and\\ C,\quad B\quad \& \quad D\quad are\quad collinear.\\ \angle AED=z,\quad \angle APC=95^{ o }\quad \& \quad \angle PCB=40^{ o }.\\ To\quad find\quad out-\\ z=?\\ solution-\\ \angle APC+\angle ABC={ 180 }^{ o }\quad (since\quad the\quad sum\quad of\quad the\quad opposite\quad angles\quad \\ of\quad a\quad cyclic\quad quadrilateral\quad is\quad { 180 }^{ o }).\\ \Longrightarrow \angle ABC={ 180 }^{ o }-\angle APC={ 180 }^{ o }-95^{ o }=85^{ o }.\\ Again\quad \angle ABC+\angle ABD={ 180 }^{ o }(linear\quad pair).\\ \therefore \quad \angle ABD={ 180 }^{ o }-\angle ABC={ 180 }^{ o }-85^{ o }=95^{ o }.\\ Now\quad \angle ABD+\angle AED={ 180 }^{ o }\quad (since\quad the\quad sum\quad of\quad the\quad opposite\quad angles\quad \\ of\quad a\quad cyclic\quad quadrilateral\quad is\quad { 180 }^{ o }).\\ \therefore \quad \angle AED=z={ 180 }^{ o }-\angle ABD={ 180 }^{ o }-95^{ o }=85^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.$$

$$ Given-\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Its\quad diagonals\quad AB\quad \& \quad CD\quad intersect\quad at\quad E.\\ \angle DBC={ 70 }^{ o },\quad \angle BAC={ 30 }^{ o }.\\ To\quad find\quad out-\\ \angle BCD=?\\ Also,\quad if\quad AB=BC\quad then\quad \angle ECD=?\\ Solution-\\ BC\quad subtends\quad \angle BDC\quad \& \quad \angle BAC\quad to\quad the\quad circumference\quad \\ of\quad the\quad given\quad circle\quad at\quad D\quad \& \quad A\quad respectively.\\ \therefore \quad \angle BDC=\angle BAC={ 30 }^{ o }\quad [since\quad the\quad angles,\quad subtended\quad by\quad \\ a\quad chord\quad of\quad a\quad circle\quad to\quad different\quad points\quad of\quad the\quad circumferece\quad \\ of\quad the\quad same\quad circle,\quad are\quad equal].\\ \therefore \quad \angle BCD=180^{ o }-(\angle DBC+\angle BDC)\quad \left( angle\quad sum\quad property\quad of\quad triangles \right) \\ \Longrightarrow \angle BCD=180^{ o }-({ 70 }^{ o }+{ 30 }^{ o })={ 80 }^{ o }.\\ Again,\quad in\quad \Delta ABC\quad we\quad have\quad AB=BC.\\ i.e\quad \Delta ABC\quad is\quad isosceles\quad with\quad AC\quad as\quad base.\\ \therefore \quad \angle BAC=\angle BCA={ 30 }^{ o }.\\ So\quad \angle ECD=\angle BCD-\angle BCA=\quad { 80 }^{ o }-{ 30 }^{ o }={ 50 }^{ o }.\\ \therefore \quad \angle BCD\quad \& \quad \angle ECD\quad are\quad respectively \quad { 80 }^{ o }\quad \& \quad { 50 }^{ o }.\\ Hence,\quad option\quad C \quad is \quad correct.  $$