Question

the diagram shows $\triangle abc$ and $\overrightarrow{bd}$. what is the measure of $\angle acd$?

Explanation:

Step1: Find ∠ACB in △ABC

In right - triangle \(ABC\), \(\angle B = 90^{\circ}\) and \(\angle A=43^{\circ}\). We know that the sum of the interior angles of a triangle is \(180^{\circ}\). So, \(\angle ACB=180^{\circ}-\angle A - \angle B\).
Substitute \(\angle A = 43^{\circ}\) and \(\angle B = 90^{\circ}\) into the formula: \(\angle ACB=180^{\circ}-43^{\circ}-90^{\circ}=47^{\circ}\).

Step2: Find ∠ACD

Since \(\angle ACB\) and \(\angle ACD\) are supplementary angles (they form a linear pair, so their sum is \(180^{\circ}\)), we have \(\angle ACD = 180^{\circ}-\angle ACB\).
Substitute \(\angle ACB = 47^{\circ}\) into the formula: \(\angle ACD=180^{\circ}-47^{\circ}=133^{\circ}\).

Answer:

\(133^{\circ}\)