To solve part e of the given problem, we need to find the value of angle $x$ in the triangle $ABC$. The diagram gives the following information:

• $\angle BAC = x$
• $\angle ABC = 40^\circ$
• $\angle ACD = 65^\circ$

Steps to Solve:

1. Identify given angles:

   • $\angle ABC = 40^\circ$
   • $\angle ACD = 65^\circ$
   • $\angle BAC = x$

2. Determine the relationship between the angles: Since $BC$ is a straight line, the angles at point $C$ form a linear pair. Hence: $\angle BCD + \angle ACD = 180^\circ$

3. Find ( \angle BCD: Given $\angle ACD = 65^\circ$, we can calculate $\angle BCD$ as: $\angle BCD = 180^\circ - 65^\circ = 115^\circ$

4. Sum of angles in triangle $ABC$: The sum of the angles in any triangle is always $180^\circ$. Hence for $\triangle ABC$: $\angle BAC + \angle ABC + \angle BCA = 180^\circ$ Substituting the values we know: $x + 40^\circ + y = 180^\circ$ where $y$ is the unknown angle at $C$.

5. Calculate $y$: Since angle $y = \angle BCA = \angle BCD = 115^\circ$:

4. Solve for x:

( y + 40+ x= 180 \ 115 +40 +x= 180 \

 X= 180 -155= 25

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