Question

1. a triangle abc, angle a is 35° and angle b is 20°. select all triangles which are similar to triangle abc.

(a) triangle def where angle d is 35° and angle e is 20°
(b) triangle ghi where angle g is 35° and angle j is 30°
(c) triangle jkl where angle j is 35° and angle l is 125°
(d) triangle mno where angle n is 20° and angle o is 125°
(e) triangle pqr where angle q is 20° and angle r is 30°

Explanation:

Step1: Calculate angle \(C\) in \(\triangle ABC\)

In \(\triangle ABC\), using the angle - sum property of a triangle (\(A + B+C=180^{\circ}\)). Given \(A = 35^{\circ}\) and \(B = 20^{\circ}\), then \(C=180-(35 + 20)=125^{\circ}\)

Step2: Check each triangle

• For \(\triangle DEF\):

Given \(D = 35^{\circ}\), \(E = 20^{\circ}\), then \(F=180-(35 + 20)=125^{\circ}\). Since \(\angle A=\angle D = 35^{\circ}\), \(\angle B=\angle E = 20^{\circ}\), \(\angle C=\angle F = 125^{\circ}\), by the AA (angle - angle) similarity criterion (if two angles of one triangle are equal to two angles of another triangle, the triangles are similar), \(\triangle ABC\sim\triangle DEF\)

• For \(\triangle GHI\):

Given \(G = 35^{\circ}\), \(I = 30^{\circ}\), then \(H=180-(35 + 30)=115^{\circ}\). Since the angles are not equal to the angles of \(\triangle ABC\), \(\triangle ABC\) and \(\triangle GHI\) are not similar

• For \(\triangle JKL\):

Given \(J = 35^{\circ}\), \(L = 125^{\circ}\), then \(K=180-(35 + 125)=20^{\circ}\). But the order of angles (if we consider correspondence) is not the same as in \(\triangle ABC\) (in \(\triangle ABC\), \(A = 35^{\circ}\), \(B = 20^{\circ}\), \(C = 125^{\circ}\)). If we assume correspondence based on angle measures, \(\triangle ABC\) and \(\triangle JKL\) are similar (by AA, since \(A = J=35^{\circ}\), \(B = K = 20^{\circ}\), \(C = L=125^{\circ}\))

• For \(\triangle MNO\):

Given \(N = 20^{\circ}\), \(O = 125^{\circ}\), then \(M=180-(20 + 125)=35^{\circ}\). Since \(\angle A=\angle M = 35^{\circ}\), \(\angle B=\angle N = 20^{\circ}\), \(\angle C=\angle O = 125^{\circ}\), by the AA similarity criterion, \(\triangle ABC\sim\triangle MNO\)

• For \(\triangle PQR\):

Given \(Q = 20^{\circ}\), \(R = 30^{\circ}\), then \(P=180-(20 + 30)=130^{\circ}\). Since the angles are not equal to the angles of \(\triangle ABC\), \(\triangle ABC\) and \(\triangle PQR\) are not similar

So, the triangles similar to \(\triangle ABC\) are \(\triangle DEF\) and \(\triangle MNO\)

Answer:

A. Triangle \(DEF\) where angle \(D\) is \(35^{\circ}\) and angle \(E\) is \(20^{\circ}\); D. Triangle \(MNO\) where angle \(N\) is \(20^{\circ}\) and angle \(O\) is \(125^{\circ}\)