Question

Find the value of x, y, z in the following diagram:

Pls send the answer step by step it's urgent....I will mark the answer branliest! ​

Answer 1

Question Point.

If two angles in an isosceles triangle are equal, both angles are called base angles. The last remaining angle is called a vertex angle.

In a triangle, an external angle is equal to the sum of two angles that are not neighboring. [1]

Solution.

$\angle x=32^{\circ}$ (A base angle.)

$\angle y=64^{\circ}$ (An external angle of a triangle.)

$\angle z=52^{\circ}$ (A vertex angle.)

Reason.

[1] Let two angles of a triangle be $\angle x,\angle y$. The remaining angle is $180^{\circ}-(\angle x+\angle y)$. Since external and internal angle sum to 180°, the external angle is the sum of two angles $\angle x+\angle y$.

Answer 2

$\Large{\textsf{\textbf{\underline{\underline{To\:do\::}}}}}$

1. Value of x.
2. Value of y.
3. Value of z.

$\Large{\textsf{\textbf{\underline{\underline{Solution\::}}}}}$

As we know that,

✯ If two sides of a triangle are equal, then their corresponding angles are also equal to each other.

In ∆ABC,

$\dashrightarrow\:{\tt{BC\:=\:AC}}~~~(Given)$

Thus,

✔ Their corresponding angles are also equal.

$\dashrightarrow\:\tt{\angle{BAC}\:=\:\angle{ABC}\:}$

According to the question,

• $\rm{\angle{BAC}\:=\:32^{\circ}\:}$

$\dashrightarrow\:\tt{\angle{ABC}\:=\:x\:=\:32^{\circ}\:}$

$\dashrightarrow\:\bf\pink{x\:=\:32^{\circ}\:}$

Again,

In ∆ABC,

➠ $\tt{\angle{BAC}\:+\:\angle{ABC}\:+\:\angle{ACB}\:=\:180^{\circ}\:}$

➠ $\tt{32^{\circ}\:+\:32^{\circ}\:+\:\angle{ACB}\:=\:180^{\circ}\:}$

➠ $\tt{64^{\circ}\:+\:\angle{ACB}\:=\:180^{\circ}\:}$

➠ $\tt{\angle{ACB}\:=\:180^{\circ}\:-\:64^{\circ}\:}$

➠ $\bf{\angle{ACB}\:=\:116^{\circ}\:}$

As we know that,

✔ Angle made by a straight line is always 180°.

$\dashrightarrow\:\tt{\angle{ACB}\:+\:y\:=\:180^{\circ}\:}$

$\dashrightarrow\:\tt{116^{\circ}\:+\:y\:=\:180^{\circ}\:}$

$\dashrightarrow\:\tt{y\:=\:180^{\circ}\:-\:116^{\circ}\:}$

$\dashrightarrow\:{\bf{\pink{y\:=\:64^{\circ}\:}}}$

Now,

In ∆BCD,

$\longrightarrow\:{\tt{BC\:=\:BD}}~~~(Given)$

$\longrightarrow\:\tt{\angle{BDC}\:=\:y\:}$

$\longrightarrow\:\tt{\angle{BDC}\:=\:64^{\circ}\:}$

Again,

In ∆BCD,

$\dashrightarrow\:\tt{\angle{BDC}\:+\:y\:+\:z\:=\:180^{\circ}\:}$

$\dashrightarrow\:\tt{64^{\circ}\:+\:64^{\circ}\:+\:z\:=\:180^{\circ}\:}$

$\dashrightarrow\:\tt{128^{\circ}\:+\:z\:=\:180^{\circ}\:}$

$\dashrightarrow\:\tt{z\:=\:180^{\circ}\:-\:128^{\circ}\:}$

$\dashrightarrow\:{\bf{\pink{z\:=\:52^{\circ}\:}}}$