Question

what kind of triangle is bcd?
a right
b acute
c obtuse
d equiangular

Explanation:

Step1: Find the measure of ∠BDC

Since ∠BDC and the \(130^{\circ}\) angle are supplementary (they form a linear pair), we use the formula \(\angle BDC = 180^{\circ}-130^{\circ}\).
\(\angle BDC=50^{\circ}\)

Step2: Find the measure of ∠DBC

In \(\triangle BCD\), we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(\angle DBC = x\). Using the angle - sum formula \(\angle DBC+\angle BDC+\angle BCD = 180^{\circ}\), and \(\angle BCD = 70^{\circ}\), \(\angle BDC = 50^{\circ}\).
\(x+50^{\circ}+70^{\circ}=180^{\circ}\), then \(x=\angle DBC=180^{\circ}-(50^{\circ} + 70^{\circ})=60^{\circ}\)

Step3: Find the measure of ∠1

Since \(\angle1\) and \(\angle2\) are adjacent angles and \(\angle DBC=\angle1+\angle2\), and we assume \(\angle1=\angle2\) (from the property of the figure, if it is a bisected angle - like situation, but actually, in a non - ambiguous case, we can also use the fact that \(\angle DBC\) is composed of two angles. Wait, no, actually, if we consider the triangle properties again. Wait, another approach:
We know that \(\angle BDC = 50^{\circ}\), \(\angle BCD=70^{\circ}\), so \(\angle DBC = 60^{\circ}\).
If we assume the line is a straight line at \(B\) (the vertical line), and if we consider the triangle \(BCD\) again. Wait, no, let's re - check.
Wait, in \(\triangle BCD\), angles: \(\angle BDC = 50^{\circ}\), \(\angle BCD=70^{\circ}\), so \(\angle DBC=60^{\circ}\).
Now, if we assume that the triangle \(BCD\) is the one in question. Since all angles \(\angle BDC = 50^{\circ}\), \(\angle BCD=70^{\circ}\), \(\angle DBC=60^{\circ}\) are less than \(90^{\circ}\), the triangle is acute.

Answer:

B. acute