Question
The vertical angle of an isosceles triangle is $100^\circ $. Find its base angles.
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Answer
Consider an isosceles $\triangle\text{ABC}$ such that $\text{AB}=\text{AC}$
Given that vertical angle A is $100^\circ $
To find the base angles Since $\triangle\text{ABC}$ is isosceles $\angle\text{B}=\angle\text{C}$
[Angles opposite to equal sides are equal] And
also, Sum of interior angles of a triangle $= 180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$(10)0^\circ+\angle\text{B}\angle\text{B}=180^\circ$
$2\angle\text{B}=180^\circ-(10)0^\circ$
$\angle\text{B}=40^\circ$$\angle\text{B}=\angle\text{C}=40^\circ$
Given that vertical angle A is $100^\circ $
To find the base angles Since $\triangle\text{ABC}$ is isosceles $\angle\text{B}=\angle\text{C}$
[Angles opposite to equal sides are equal] And
also, Sum of interior angles of a triangle $= 180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$(10)0^\circ+\angle\text{B}\angle\text{B}=180^\circ$
$2\angle\text{B}=180^\circ-(10)0^\circ$
$\angle\text{B}=40^\circ$$\angle\text{B}=\angle\text{C}=40^\circ$
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