Calculation of Angles in Triangles
To find the value of \( x \), we need to look at the triangle \( ABC \). Since we know two angles in the triangle, \( 15^\circ \) (angle \( A \)) and \( 45^\circ \) (angle \( B \)), we can find angle \( C \) because the sum of the angles in a triangle is \( 180^\circ \).
Let's calculate angle \( C \):
\( C = 180^\circ - A - B = 180^\circ - 15^\circ - 45^\circ = 120^\circ \)
Now we can focus on triangle \( BCD \), it appears to be an isosceles triangle since \( BD = DC \). Therefore, angle \( B \) and angle \( C \) in this triangle should be equal. However, were also given that \( DE \) is congruent to \( BD \), which means triangle \( BDE \) is also isosceles, and angles \( B \) and \( E \) must be equal.
Since the large angle \( C \) of the big triangle \( ABC \) is split into two angles \( C \) and \( E \) within the triangle \( BCD \) and \( BDE \) respectively, we must have:
\( C \) (angle within triangle \( BCD \)) \( + E = 120^\circ \) (angle C of \( ABC \))
Since \( C \) (within \( BCD \)) is equal to \( B = 45^\circ \), we get:
\( 45^\circ + E = 120^\circ \)
Now we can solve for \( E \):
\( E = 120^\circ - 45^\circ = 75^\circ \)
This is the value of angle \( E \), and since \( DE \) is congruent to \( BD \), angle \( D \) in triangle \( BDE \) is also \( 75^\circ \). Our goal is to find the angle \( x \) which is the angle \( D \) in triangle \( ADE \). Since \( ADE \) is an isosceles triangle with \( AE = DE \), the angles at \( A \) and \( D \) are equal.
Therefore, angle \( x \) (which is angle \( D \) in \( ADE \)) equals \( 75^\circ \).
