Question

1. find m∠ydc. y d 15x+5 c 80° 6x+6 b

Explanation:

Step1: Identify Triangle Angle Sum

In triangle \( DBC \), the sum of angles is \( 180^\circ \). So, \( (15x + 5) + 80^\circ + (6x + 6) = 180^\circ \)? Wait, no—wait, \( \angle YDC \) is supplementary to the angle at \( D \) in the triangle? Wait, no, actually, \( \angle YDC \) and the angle at \( D \) in triangle \( DBC \) are supplementary? Wait, no, let's correct. Wait, the angle at \( D \) in triangle \( DBC \) is adjacent to \( \angle YDC \), so they are supplementary? Wait, no, looking at the diagram, \( YD \) is a straight line with \( D \) connected to \( C \) and \( B \). Wait, actually, in triangle \( DBC \), the angles are \( \angle DBC = 6x + 6 \), \( \angle BCD = 80^\circ \), and \( \angle BDC \) (the angle at \( D \) in the triangle) is such that \( \angle YDC \) is \( 15x + 5 \), and \( \angle YDC \) and \( \angle BDC \) are supplementary? Wait, no, maybe \( DC \) is parallel to something? No, better to use the exterior angle theorem or triangle angle sum. Wait, actually, \( \angle YDC \) is an exterior angle? Wait, no, let's check the triangle. Wait, the sum of angles in triangle \( DBC \) is \( 180^\circ \), so \( \angle BDC + 80^\circ + (6x + 6) = 180^\circ \). But \( \angle YDC \) is \( 15x + 5 \), and \( \angle YDC + \angle BDC = 180^\circ \) (linear pair). So first, find \( x \) by using the triangle angle sum. Let's denote \( \angle BDC = 180^\circ - (15x + 5) \). Then in triangle \( DBC \): \( (180 - 15x - 5) + 80 + 6x + 6 = 180 \). Simplify: \( 175 - 15x + 80 + 6x + 6 = 180 \) → \( 261 - 9x = 180 \) → \( -9x = 180 - 261 = -81 \) → \( x = 9 \). Wait, no, that's complicated. Wait, maybe I made a mistake. Wait, actually, \( \angle YDC \) is equal to the sum of the two remote interior angles? Wait, no, \( \angle YDC \) is an exterior angle to triangle \( DBC \) at \( D \). So by exterior angle theorem, \( \angle YDC = \angle DBC + \angle BCD \). So \( 15x + 5 = (6x + 6) + 80 \). Ah, that's the exterior angle theorem! Yes, because \( \angle YDC \) is an exterior angle for triangle \( DBC \), so it equals the sum of the two non-adjacent interior angles. So:

Step2: Solve for \( x \) using Exterior Angle Theorem

Exterior Angle Theorem: \( \angle YDC = \angle DBC + \angle BCD \)
So \( 15x + 5 = (6x + 6) + 80 \)
Simplify right side: \( 6x + 86 \)
Equation: \( 15x + 5 = 6x + 86 \)
Subtract \( 6x \) from both sides: \( 9x + 5 = 86 \)
Subtract 5: \( 9x = 81 \)
Divide by 9: \( x = 9 \)

Step3: Find \( m\angle YDC \)

Substitute \( x = 9 \) into \( 15x + 5 \):
\( 15(9) + 5 = 135 + 5 = 140^\circ \)? Wait, no, that can't be, because \( 80 + 6x + 6 \) would be \( 80 + 54 + 6 = 140 \), and \( 15x + 5 = 140 \), so supplementary? Wait, no, I think I messed up the exterior angle. Wait, actually, \( \angle YDC \) and the angle at \( D \) in the triangle are supplementary. Wait, let's re-examine. The triangle has angles: \( \angle BCD = 80^\circ \), \( \angle DBC = 6x + 6 \), and \( \angle BDC \) (the angle at \( D \) inside the triangle). Then \( \angle YDC \) is adjacent to \( \angle BDC \), so \( \angle YDC + \angle BDC = 180^\circ \). And in the triangle, \( \angle BDC + 80^\circ + (6x + 6) = 180^\circ \), so \( \angle BDC = 180 - 80 - 6x - 6 = 94 - 6x \). Then \( \angle YDC = 180 - (94 - 6x) = 86 + 6x \). But \( \angle YDC \) is given as \( 15x + 5 \). So set \( 15x + 5 = 86 + 6x \). Then \( 15x - 6x = 86 - 5 \) → \( 9x = 81 \) → \( x = 9 \). Then \( \angle YDC = 15(9) + 5 = 135 + 5 = 140^\circ \). Wait, but let's check the triangle angles. \( \angle DBC = 6(9) + 6 = 60^\circ \). Then \( \angle BDC = 180 - 80 - 60 = 40^\circ \). Then \( \angle YDC + \angle BDC = 140 + 40 = 180^\circ \), which is correct (linear pair). So that works.

Answer:

\( 140^\circ \)