# Thread: triangle inscribed in a circle...

1. ## triangle inscribed in a circle...

Triangle ABC is inscribed in a circle as shown in the figure. AX,BY and CZ bisect A,BAnd C respectively. If BAC= 56 degrees, ABC=64 degrees. Find the angles of XYZ. A,Y,C,X,Z in clockwise.

2. Hello, snigdha!

Triangle $ABC$ is inscribed in a circle as shown in the figure.

$AX,BY, CZ$ bisect $\angle A, \angle B, \angle C$, respectively.

If $\angle BAC= 56^o,\;\angle ABC= 64^o,$

find the angles of $\Delta XYZ\/\text{ and }\,\angle C$
Thereom: An inscribed angle is measured by one-half its intercepted arc.

Since $\angle BAC = 56^o$, then: . $\text{arc}(BXC) \,=\,112^o$

Since $\angle ABC = 64^o$, then: . $\text{arc}(AYC) \,=\,128^o$

Hence: . $\text{arc}(BZA) \;=\;360^o - 112^o - 128^o \;=\;120^o$

Therefore: . $\boxed{\angle C \:=\:60^o}$

Since $CZ$ bisects $\angle C\!:\;\angle ACZ \,=\, \angle BCZ \,=\, 30^o \quad\Rightarrow\quad \text{arc}(BZ) \,=\,\text{arc}(ZA) \,=\, 60^o$

Since $BY$ bisects $\angle B\!:\;\angle ABY \,=\, \angle CBY \,=\,32^o \quad\Rightarrow\quad \text{arc}(AY) \,=\, \text{arc}(YC) \,=\,64^o$

Hence: . $\text{arc}(ZAY) \:=\:60^o + 64^o \:=\:124^o$

. . Therefore: . $\boxed{\angle X \:=\:62^o}$

$\text{arc}(XBZ) \:=\:56^o + 60^o \:=\:116^o \quad\Rightarrow\quad \boxed{\angle Y \:=\:58^o}$

$\text{arc}(XCY) \:=\:56^o + 64^o \:=\:120^o \quad\Rightarrow\quad\boxed{ \angle Z \:=\:60^o}$