Question

m∠abc + m∠bac + m∠acb when △abc is an isosceles triangle with ab = ac
m∠qpr when m∠qrp = 30° and △pqr is an isosceles triangle with pq = qr
m∠abc when m∠bac = 70° and △abc is an isosceles triangle with ab = ac
m∠bde when m∠bac = 45° and points d and e are the midpoints of ab and bc respectively in △abc
55°
180°
45°
30°

Explanation:

Step1: Find \(m\angle ABC + m\angle BAC + m\angle ACB\)

In any triangle, the sum of interior angles is \(180^{\circ}\). So \(m\angle ABC + m\angle BAC + m\angle ACB=180^{\circ}\)

Step2: Find \(m\angle ABC\) when \(m\angle BAC = 70^{\circ}\) and \(\triangle ABC\) is isosceles with \(AB = AC\)

Since \(AB = AC\), \(\angle ABC=\angle ACB\). Let \(m\angle ABC = x\). Then \(x + x+70^{\circ}=180^{\circ}\). Solving \(2x=180^{\circ}- 70^{\circ}=110^{\circ}\), \(x = 55^{\circ}\)

Step3: Find \(m\angle QPR\) when \(m\angle QRP = 30^{\circ}\) and \(\triangle PQR\) is isosceles with \(PQ = QR\)

Since \(PQ = QR\), \(\angle QPR=\angle QRP\). So \(m\angle QPR = 30^{\circ}\)

Step4: Find \(m\angle BDE\) when \(m\angle BAC = 45^{\circ}\) and \(D\) and \(E\) are mid - points of \(AB\) and \(BC\) respectively in \(\triangle ABC\)

By the mid - point theorem (a line segment joining the mid - points of two sides of a triangle is parallel to the third side), \(DE\parallel AC\). Then \(\angle BDE=\angle BAC\) (corresponding angles). So \(m\angle BDE = 45^{\circ}\)

Answer:

\(180^{\circ}\to m\angle ABC + m\angle BAC + m\angle ACB\)

\(55^{\circ}\to m\angle ABC\) when \(m\angle BAC = 70^{\circ}\) and \(\triangle ABC\) is isosceles with \(AB = AC\)

\(45^{\circ}\to m\angle BDE\) when \(m\angle BAC = 45^{\circ}\) and \(D\) and \(E\) are mid - points of \(AB\) and \(BC\) respectively in \(\triangle ABC\)

\(30^{\circ}\to m\angle QPR\) when \(m\angle QRP = 30^{\circ}\) and \(\triangle PQR\) is isosceles with \(PQ = QR\)