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# help geometry

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In triangle ABC, AB = AC.  Find angle BAC, in degrees.

Jun 25, 2021

#1
+151
+3

since in \$\triangle ABC\$, \$\overset{-}{AB}=\overset{-}{AC}\$, it means that it is an isosceles triangle, which also means that both angles at the bottom are going to be the same

so we know that \$\angle ABC=\angle ACB=50^\circ\$

by adding the two you get \$50^\circ+50^\circ=100^\circ\$

a triangle's interior angle are equal to \$180^\circ\$

knowing that \$ \angle ABC+\angle ACB+ \angle BAC=180^\circ  \$, by plugging in what we have we get \$ 100^\circ+\angle BAC=180^\circ  \$

\$  \angle BAC=180^\circ-100^\circ   \$

\$ \boxed{ \angle BAC=80^\circ } \$

Jun 25, 2021

#1
+151
+3

since in \$\triangle ABC\$, \$\overset{-}{AB}=\overset{-}{AC}\$, it means that it is an isosceles triangle, which also means that both angles at the bottom are going to be the same

so we know that \$\angle ABC=\angle ACB=50^\circ\$

by adding the two you get \$50^\circ+50^\circ=100^\circ\$

a triangle's interior angle are equal to \$180^\circ\$

knowing that \$ \angle ABC+\angle ACB+ \angle BAC=180^\circ  \$, by plugging in what we have we get \$ 100^\circ+\angle BAC=180^\circ  \$

\$  \angle BAC=180^\circ-100^\circ   \$

\$ \boxed{ \angle BAC=80^\circ } \$