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# ABC is a right angled triangle in which $$\angle{A}$$ = $$90^\circ$$ and AB = AC, find $$\angle{B}$$ and $$\angle{C}$$.

In $$\triangle{ABC}$$, we have,

AB = AC ...(Given)
$$\angle{B}$$ = $$\angle{C}$$ ...(i)($$\because$$ angles opposite to equal sides are equal)

Now, we know that,

$$\angle{A}$$ + $$\angle{B}$$ + $$\angle{C}$$ = $$180^\circ$$
$$\Rightarrow$$ $$90^\circ$$ + $$\angle{B}$$ + $$\angle{C}$$ = $$180^\circ$$ ...(given)
$$\Rightarrow$$ $$90^\circ$$ + $$\angle{B}$$ + $$\angle{B}$$ = $$180^\circ$$ ...(from(i))
$$\Rightarrow$$ 2 $$\angle{B}$$ = $$180^\circ$$ - $$90^\circ$$
$$\Rightarrow$$ 2 $$\angle{B}$$ = $$90^\circ$$
$$\Rightarrow$$ $$\angle{B}$$ = $$45^\circ$$
$$\therefore$$ $$\angle{C}$$ = $$45^\circ$$, too.