Question

in the diagram below, m∠cih = 99° and m∠bgd = 36°. find m∠ihg.
step angle reason
1 m∠cih = 99° given
2 m∠bgd = 36° given
try m∠ yi = □° select a reason

Explanation:

Step1: Identify corresponding angles or transversal relationships

Assuming lines are parallel (or using the exterior angle theorem or triangle angle sum), we know that \( \angle BGD \) and the angle at \( G \) related to \( \angle IHG \) might be alternate interior or corresponding, but more likely, we use the fact that in a triangle or linear pair, but here, let's consider the sum of angles. Wait, actually, \( \angle CIH = 99^\circ \), so its supplementary angle (linear pair) at \( I \) is \( 180^\circ - 99^\circ = 81^\circ \). Then, if \( \angle BGD = 36^\circ \), and we have a triangle or a transversal, but actually, the key is that \( \angle IHG \) can be found by \( 180^\circ - 99^\circ - 36^\circ \)? Wait, no, maybe using the fact that \( \angle CIH \) is an exterior angle or something. Wait, let's re-examine.

Wait, the diagram: points \( D, G, I, C \) with \( D-G-I-C \)? No, \( D \) to \( G \) to \( I \) to \( C \)? Wait, the angle at \( I \) is \( 99^\circ \) ( \( \angle CIH \) ), and at \( G \) is \( 36^\circ \) ( \( \angle BGD \) ). Let's assume that \( \angle IHG \) is in a triangle where we have two angles: the angle supplementary to \( 99^\circ \) (which is \( 81^\circ \)) and \( 36^\circ \), so the third angle \( \angle IHG = 180^\circ - 81^\circ - 36^\circ = 63^\circ \)? Wait, no, maybe I messed up. Wait, actually, \( \angle CIH = 99^\circ \), so the adjacent angle (linear pair) is \( 180 - 99 = 81^\circ \). Then, \( \angle BGD = 36^\circ \), and if those are two angles in a triangle, then \( \angle IHG = 180 - 81 - 36 = 63^\circ \)? Wait, no, maybe the correct approach is:

Wait, \( \angle CIH = 99^\circ \), so the angle inside the triangle at \( I \) is \( 180 - 99 = 81^\circ \). \( \angle BGD = 36^\circ \), which is equal to the angle at \( G \) (alternate interior or corresponding). Then, in triangle \( IHG \), the sum of angles is \( 180^\circ \), so \( \angle IHG = 180 - 81 - 36 = 63^\circ \). Wait, but maybe it's simpler: \( \angle IHG = 180 - 99 - 36 = 45^\circ \)? No, that's wrong. Wait, no, let's do it properly.

Wait, \( \angle CIH = 99^\circ \), so the angle at \( I \) (interior to the triangle) is \( 180 - 99 = 81^\circ \). \( \angle BGD = 36^\circ \), which is the angle at \( G \) (interior to the triangle). Then, \( \angle IHG = 180 - 81 - 36 = 63^\circ \). Wait, but maybe the correct answer is \( 180 - 99 - 36 = 45 \)? No, 180 - 99 is 81, 81 - 36 is 45? Wait, 81 - 36 is 45? No, 81 - 36 = 45? Wait, 80 - 36 is 44, plus 1 is 45. Yes. Wait, I miscalculated earlier. 81 - 36 = 45. So \( \angle IHG = 45^\circ \)? Wait, no, maybe the correct approach is:

Wait, \( \angle CIH = 99^\circ \), so the angle adjacent (linear pair) is \( 180 - 99 = 81^\circ \). Then, \( \angle BGD = 36^\circ \), and since those are two angles in a triangle, the third angle \( \angle IHG = 180 - 81 - 36 = 63^\circ \)? Wait, I'm confused. Wait, let's use the exterior angle theorem. The exterior angle \( \angle CIH = 99^\circ \) is equal to the sum of the two remote interior angles, which would be \( \angle BGD = 36^\circ \) and \( \angle IHG \). Wait, that's the key! Exterior angle theorem: the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. So \( \angle CIH = \angle BGD + \angle IHG \). Therefore, \( 99^\circ = 36^\circ + \angle IHG \).

Step2: Solve for \( \angle IHG \)

Using the exterior angle theorem: \( \angle IHG = \angle CIH - \angle BGD \)
\( \angle IHG = 99^\circ - 36^\circ = 63^\circ \)

Wait, that makes sense. Because \( \angle CIH \) is an exterior angle to the triangle with interior angles \( \angle BGD \) and \( \angle IHG \), so by exterior angle theorem, exterior angle = sum of two remote interior angles. So \( 99 = 36 + \angle IHG \), so \( \angle IHG = 99 - 36 = 63 \).

Answer:

\( \boldsymbol{63^\circ} \)