R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: EXERCISE 12C

In the given figure, $\mathrm{AB}$ is a diameter of a circle with centre $\mathrm{O}$ and $\mathrm{DO}\parallel \mathrm{CB}$. If $\angle \mathrm{BCD}=120°$ and $\angle \mathrm{ABD}=k°$, then find the value of $k$.

In the given figure, $\mathrm{AB}$ is a diameter of a circle with center $\mathrm{O}$ and $\mathrm{DO}\parallel \mathrm{CB}$. If $\angle \mathrm{BCD}=120°$ calculate $\angle \mathrm{CBD}$.

Two chords $\mathrm{AB}$ and $\mathrm{CD}$ of a circle intersect each other at $\mathrm{P}$ outside the circle. $\mathrm{AB}=6 \mathrm{cm}, \mathrm{BP}=2 \mathrm{cm}$ and $\mathrm{PD}=2.5 \mathrm{cm}$. If the length of $\mathrm{CD}$ is $k \mathrm{cm}$, then find the value of $k$. 

In the given figure, $O$ is the centre of a circle. If $\angle AOD=140°$, $\angle CAB=50°$ and $\angle EDB=k°$, then find the value of $k$.

In the given figure, $O$ is the centre of a circle. If $\angle AOD=140°$ and $\angle CAB=50°$, calculate $\angle EBD$.

In the given figure, $\angle BAD=75°, \angle DCF=x°$ and $\angle DEF=y°$. Find the values of $y-x$.

On a common hypotenuse $AB,$ two right triangles $ACB$ and $ADB$ are situated on opposite sides. Prove that $\angle BAC=\angle BDC.$

$ABCD$ is a quadrilateral such that $A$ is the center of the circle passing through $B, C$ and $D.$ Prove that $\angle \mathrm{CBD}+\angle \mathrm{CDB}=\frac{1}{2}\angle \mathrm{BAD}$