Question

what is ( mangle cad )?
options: ( 20^circ ), ( 80^circ ), ( 100^circ ), ( 110^circ )
(image shows a triangle ( abc ) with ( angle b = 20^circ ), ( angle c = 80^circ ), and a straight line ( dab ) with ( a ) as the vertex.)

Explanation:

Step1: Find ∠BAC in △ABC

In triangle \(ABC\), the sum of interior angles is \(180^\circ\). Given \(\angle B = 20^\circ\) and \(\angle C = 90^\circ\) (assuming the \(90^\circ\) from the diagram, though labeled as \(90^\circ\) - likely a typo but contextually a right angle or given angle). So \(\angle BAC=180^\circ - 20^\circ - 90^\circ = 70^\circ\)? Wait, no, wait the diagram: Wait, the triangle has \(\angle B = 20^\circ\), \(\angle C = 90^\circ\)? Wait, no, looking at the diagram, point \(C\) has an angle, maybe \(\angle C = 90^\circ\)? Wait, no, the diagram shows \(\angle B = 20^\circ\), and then we need to find \(\angle CAD\). Wait, \(D - A - B\) is a straight line, so \(\angle DAB = 180^\circ\). So first, find \(\angle BAC\) in triangle \(ABC\).

Wait, triangle \(ABC\): angles at \(B\) is \(20^\circ\), angle at \(C\) – wait, the diagram has \(C\) with a \(90^\circ\)? Wait, no, the user's diagram: let's re-express. The triangle \(ABC\) has \(\angle B = 20^\circ\), \(\angle C = 90^\circ\)? Wait, no, maybe \(\angle C = 90^\circ\) is a typo, but looking at the options, let's recalculate. Wait, in triangle \(ABC\), sum of angles: \(\angle B + \angle C + \angle BAC = 180^\circ\). If \(\angle B = 20^\circ\), \(\angle C = 90^\circ\) (as per diagram, the angle at \(C\) is \(90^\circ\) – maybe a right triangle). Then \(\angle BAC = 180 - 20 - 90 = 70^\circ\). Then, since \(D - A - B\) is a straight line, \(\angle CAD + \angle BAC = 180^\circ\)? Wait, no, \(D\), \(A\), \(B\) are colinear, so \(\angle DAB = 180^\circ\), so \(\angle CAD + \angle BAC = 180^\circ\)? Wait, no, \(A\) is the vertex, so \(D - A - B\) is a straight line, so \(\angle DAB = 180^\circ\), and \(\angle BAC\) is inside the triangle, so \(\angle CAD = 180^\circ - \angle BAC\). Wait, but if \(\angle BAC = 70^\circ\), then \(\angle CAD = 180 - 70 = 110^\circ\)? Wait, but the options include \(110^\circ\). Wait, maybe I made a mistake. Wait, let's check again.

Wait, maybe the angle at \(C\) is \(90^\circ\)? No, the diagram shows angle at \(C\) as \(90^\circ\) (the symbol), but maybe it's \(90^\circ\). Wait, \(\angle B = 20^\circ\), \(\angle C = 90^\circ\), so \(\angle BAC = 70^\circ\). Then, since \(D\), \(A\), \(B\) are straight, \(\angle CAD = 180^\circ - \angle BAC = 180 - 70 = 110^\circ\). Wait, but let's confirm.

Wait, another approach: exterior angle theorem. The exterior angle at \(A\) (∠CAD) is equal to the sum of the two non-adjacent interior angles of triangle \(ABC\). So \(\angle CAD = \angle B + \angle C\). If \(\angle B = 20^\circ\) and \(\angle C = 90^\circ\), then \(\angle CAD = 20 + 90 = 110^\circ\). Yes, that matches the exterior angle theorem. So that's the key.

Step2: Apply Exterior Angle Theorem

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two remote interior angles. Here, \(\angle CAD\) is an exterior angle to triangle \(ABC\) at vertex \(A\), so \(\angle CAD = \angle B + \angle C\). Given \(\angle B = 20^\circ\) and \(\angle C = 90^\circ\) (from the diagram, the right angle at \(C\)), so \(\angle CAD = 20^\circ + 90^\circ = 110^\circ\).

Answer:

\(110^\circ\) (corresponding to the option with \(110^\circ\))