Question
In Fig. $\angle\text{ADC}=130^\circ$ and chord $BC =$ chord $BE.$ Find $\angle\text{CBE}.$
✓
Answer
In the given figure, we have $\angle\text{ADC}=130^\circ$ and chord $BC = BE.$
We have to find $\angle\text{CBE}.$
Since $ABCD$ is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral are supplementary. $\therefore\angle\text{D}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\angle\text{OBC}=50^\circ....(1)$ In
$\triangle\text{OBC}$ and $\triangle\text{OBE},$
we have $BC = BE [$Given$] OC = OE [$Radii of same circle$] OB = OB [$Common side$]$
$\therefore\triangle\text{OBC}\cong\triangle\text{OBE} [$By $SSS$ cong. Rule$]$
$\angle\text{OBC}+\angle\text{OBE}=50^\circ$
$\big[$By $C.P.C.T.$ and by $(1) \angle\text{OBC}=50^\circ\big]$
$\therefore\angle\text{OBC}+\angle\text{OBE}=50^\circ+50^\circ=100^\circ$
Hence, $\angle\text{CBE}=100^\circ$
We have to find $\angle\text{CBE}.$
Since $ABCD$ is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral are supplementary. $\therefore\angle\text{D}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\angle\text{OBC}=50^\circ....(1)$ In
$\triangle\text{OBC}$ and $\triangle\text{OBE},$
we have $BC = BE [$Given$] OC = OE [$Radii of same circle$] OB = OB [$Common side$]$
$\therefore\triangle\text{OBC}\cong\triangle\text{OBE} [$By $SSS$ cong. Rule$]$
$\angle\text{OBC}+\angle\text{OBE}=50^\circ$
$\big[$By $C.P.C.T.$ and by $(1) \angle\text{OBC}=50^\circ\big]$
$\therefore\angle\text{OBC}+\angle\text{OBE}=50^\circ+50^\circ=100^\circ$
Hence, $\angle\text{CBE}=100^\circ$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If both $x+1$ and $x-1$ are factors of $a x^3+x^2-2 x+b$, find the values of $a$ and $b$.
→If measures opposite angles of a parallelogram are $(60 - x)^\circ$ and $(3x - 4)^\circ$, then find the measures of angles of the parallelogram.
→In Fig. $X$ and $Y$ are the mid-points of $AC$ and $AB$ respectively, $QP || BC$ and $CYQ$ and $BXP$ are straight lines. Prove that $ar\ (ABP) = ar\ (ACQ).$
→A cylindrical water tank of diameter $1.4\ m$ and height $2.1\ m$ is being fed by a pipe of diameter $3.5\ cm$ through which water flows at the rate of $2$ metre per second. In how much time the tank will be filled?
→If $\text{x}=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$ and $\text{y}=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$ then find the value of $x^2 + y^2$.
→Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.
In a $\triangle\text{ABC}, E$ and $F$ are the mid-points of $AC$ and $AB$ respectively. The altitude $AP$ to $BC$ intersects $FE$ at $Q.$ Prove that $AQ = QP.$
→If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.
A river $3m$ deep and $40\ m$ wide is flowing at the rate of $2\ km$ per hour. How much water will fall into the sea in a minute?
→