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if m∠b is 54° what is the m∠cdb?

Question

if m∠b is 54° what is the m∠cdb?

Explanation:

Step1: Identify the triangle type

Since \( BE = BC \) and \( ED = CD \), and \( BD = BD \) (common side), \(\triangle BED\cong\triangle BCD\) (by SSS - Side - Side - Side congruence criterion). Also, in \(\triangle BCD\), \( BC = CD\) (from the congruence and given equal - length markings), so \(\triangle BCD\) is an isosceles triangle.

Step2: Use the angle - sum property of a triangle

In \(\triangle BCD\), let \(m\angle CDB=m\angle CBD = x\) (angles opposite to equal sides \(BC\) and \(CD\) in \(\triangle BCD\)). We know that \(m\angle B = 54^{\circ}\).
The sum of angles in \(\triangle BCD\) is \(180^{\circ}\). Using the formula \(m\angle B+m\angle CDB + m\angle BCD=180^{\circ}\), but since \(m\angle CDB=m\angle CBD\) (isosceles triangle property), and we can also use the fact that for \(\triangle BCD\), \(m\angle CDB=\frac{180 - m\angle B}{2}\)

Step3: Calculate the angle

Substitute \(m\angle B = 54^{\circ}\) into the formula \(x=\frac{180 - 54}{2}\).

$$x=\frac{126}{2}=36^{\circ}$$

Answer:

\(36^{\circ}\)