In the given figure, A B=D B and A C=D C. If ABD=58^ , DBC=(2 x-4)^ , ACB=y+15^ and DCB=63^ ; find the values of x and y .

Given: In the figure AB=DB, AC=DC, ABD=58^ , DBC=(2 x-4)^ , ACB=(y+15)^ and DCB=63^ We need to find the values of x and y. In A B C and D B C AB=DB [ Given ] AC=DC [ Given ] BC=BC [ common ] By Side-SIde-Side criterion of congruence, we have, ABC DBC The corresponding parts of the congruent triangles are congruent. ABC=DCB [ c. p. c.t ] y^ +15^ =63^ y^ =63^ -15^ y^ =48^ and ABC= DBC ... [c.p.c.t ] But, DBC=(2 x-4)^ We have ABC+ DBC= ABD (2 x-4)^ +(2 x-4)^ =58^ 4 x-8^ =58^ 4 x=58^ +8^ 4 x=66^ x=66^ /(4)^ x=16.5^ Thus the values of x and y are : x=16.5^ and y=48^

In the given figure, A B=D B and A C=D C. If ABD=58^ , DBC=(2 x-4…

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Question

In the given figure, $A B=D B$ and $A C=D C$.

If $\angle \mathrm{ABD}=58^{\circ}, \angle \mathrm{DBC}=(2 \mathrm{x}-4)^{\circ},\angle \mathrm{ACB}=\mathrm{y}+15^{\circ}$ and $\angle \mathrm{DCB}=63^{\circ} $; find the values of $x$ and $y .$

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Answer

Given: In the figure $\mathrm{AB}=\mathrm{DB}, \mathrm{AC}=\mathrm{DC}, \angle \mathrm{ABD}=58^{\circ}, \angle \mathrm{DBC}=(2 \mathrm{x}-4)^{\circ}, \angle \mathrm{ACB}=(\mathrm{y}+15)^{\circ}$ and $\angle \mathrm{DCB}=63^{\circ}$
We need to find the values of $x$ and $y$.
In $\triangle A B C$ and $\triangle D B C$
$\mathrm{AB}=\mathrm{DB} \ldots[$ Given $]$
$\mathrm{AC}=\mathrm{DC} \ldots[$ Given $]$
$\mathrm{BC}=\mathrm{BC} \ldots[$ common $]$
$\therefore$ By Side$-$SIde$-$Side criterion of congruence, we have,
$\triangle \mathrm{ABC} \cong \triangle \mathrm{DBC}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \angle \mathrm{ABC}=\mathrm{DCB} \ldots[\text { c. p. c.t }]$
$ \Rightarrow \mathrm{y}^{\circ}+15^{\circ}=63^{\circ}$
$ \Rightarrow \mathrm{y}^{\circ}=63^{\circ}-15^{\circ}$
$ \Rightarrow \mathrm{y}^{\circ}=48^{\circ}$
and $\angle \mathrm{ABC}=\angle \mathrm{DBC}\dots... [c.p.c.t ]$
But, $\angle \mathrm{DBC}=(2 \mathrm{x}-4)^{\circ}$
We have $\angle \mathrm{ABC}+\angle \mathrm{DBC}=\angle \mathrm{ABD}$
$\Rightarrow(2 x-4)^{\circ}+(2 x-4)^{\circ}=58^{\circ}$
$ \Rightarrow 4 x-8^{\circ}=58^{\circ}$
$ \Rightarrow 4 x=58^{\circ}+8^{\circ}$
$ \Rightarrow 4 x=66^{\circ}$
$ \Rightarrow x=66^{\circ} /(4)^{\prime}$
$ \Rightarrow x=16.5^{\circ}$
Thus the values of $x$ and $y$ are :
$x=16.5^{\circ}$ and $y=48^{\circ}$