Chapter 7, Problem 20RP

To determine

To calculate : The value of $m\angle BDC$

Answer to Problem 20RP

The final value of $\angle BDC={115}^0$

Explanation of Solution

Given information : The following information has been given

  $\overrightarrow{AB}\cong \overrightarrow{AC}$

  $\angle DBC\cong \angle DCA$

  $m\angle A={50}^0$

Formula used : summation of the angles of any triangle is ${180}^0$

Calculation : We know that $\overrightarrow{AB}\cong \overrightarrow{AC}$ , thus, we know that

  $\angle ACB\cong \angle ABC=\frac{1}{2}({180}^0-{50}^0)={65}^0$

Let’s now assume

  $\angle DBC\cong \angle DCA=x$

Thus, we get

  $\angle DCB={65}^0-x$

Now, in triangle DBC,

  $\angle BDC={180}^0-(x+{65}^0-x)$

Thus, $\angle BDC={115}^0$