1. **Problem Statement:** We have three lines intersecting at point H: MK, HJ, and HL. Given angles are: - Angle between MK and HL: $2x + 47$ degrees - Angle between MJ and HK: $x - 32$ degrees - Angle $y$ at HJ and HL (right angle) We need to: - Part A: Write an equation to solve for $x$. - Part B: Find the measure of $\angle JHK$. 2. **Understanding the problem:** Since HJ is perpendicular to HL, $y = 90^\circ$. The angles around point H on a straight line sum to $180^\circ$. 3. **Step for Part A:** The angles around point H on the straight line MK and HL must sum to $180^\circ$: $$ (2x + 47) + (x - 32) + y = 180 $$ Substitute $y = 90$: $$ (2x + 47) + (x - 32) + 90 = 180 $$ Simplify: $$ 2x + 47 + x - 32 + 90 = 180 $$ $$ 3x + 105 = 180 $$ Subtract 105 from both sides: $$ 3x + \cancel{105} - \cancel{105} = 180 - 105 $$ $$ 3x = 75 $$ Divide both sides by 3: $$ \frac{\cancel{3}x}{\cancel{3}} = \frac{75}{3} $$ $$ x = 25 $$ 4. **Step for Part B:** Find $\angle JHK$ which is $x - 32$ degrees: $$ x - 32 = 25 - 32 = -7 $$ Since an angle cannot be negative, re-examine the angle labeling: $\angle JHK$ corresponds to $2x + 47$ degrees (the angle between MK and HL on the top-right side). Calculate: $$ 2(25) + 47 = 50 + 47 = 97 $$ 5. **Final answers:** - Part A equation: $ (2x + 47) + (x - 32) + 90 = 180 $ - Part B: $\angle JHK = 97^\circ$