Solve the following. In an isosceles triangle, the base angles are equal. The vertex angle is 40^ . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180^ ).

Let A B C be an isosceles triangle whose base angles are equal and let measure of base angle be x. Also, vertex angle is 40^ . Since, the sum of three angles of a triangle is 180^ . lr A+ B+ C=180^ 40^ +x+x=180^ 40^ +2 x=180^ which is the required equation. On transposing (+40) from LHS to RHS, we get & 2 x=180^ -40^ & 2 x=140^ On dividing both sides by 2, we get 2 x 2 = 140^ 2 =70^ x=70^ Hence, the base angles of the triangle are of measure 70°.

Solve the following. In an isosceles triangle, the base angles ar…

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Question

Solve the following.
In an isosceles triangle, the base angles are equal. The vertex angle is $40^{\circ}$. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180^{\circ}$ ).

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Answer

Let $A B C$ be an isosceles triangle whose base angles are equal and let measure of base angle be $x$. Also, vertex angle is $40^{\circ}$.
Since, the sum of three angles of a triangle is $180^{\circ}$.
$
\begin{array}{lr}
\therefore \angle A+\angle B+\angle C=180^{\circ} \\
\Rightarrow 40^{\circ}+x+x=180^{\circ} \\
\Rightarrow 40^{\circ}+2 x=180^{\circ}
\end{array}
$
which is the required equation.
On transposing (+40) from LHS to RHS, we get
$\begin{aligned} & 2 x=180^{\circ}-40^{\circ} \\ \Rightarrow \quad & 2 x=140^{\circ}\end{aligned}$
On dividing both sides by 2, we get
$\frac{2 x}{2}=\frac{140^{\circ}}{2}=70^{\circ}$
$\Rightarrow \quad x=70^{\circ}$
Hence, the base angles of the triangle are of measure 70°.