Question

ABCD is a rhombus and ΔCDE is equilateral. ∠BCD = 98°. Find x, y and z.

Answer 1

Given:

• ABCD is a rhombus, so all sides are equal and opposite angles are equal.
• ΔCDE is equilateral, so all angles there are 60°.
• ∠BCD = 98° is given inside the rhombus.
• We need to find x, y and z, which are angles formed between certain points as marked on the figure.

Stepwise calculation:

1. Since ABCD is a rhombus, adjacent angles are supplementary: ∠BCD + ∠ABC = 180°.

So if ∠BCD = 98°, then ∠ABC = 82°. 

2. Consider the triangle BCD inside rhombus ABCD:

Since ABCD is rhombus, CD = BC and both sides are equal.

∠BCD is given as 98°. Let x be the small angle between BC and CE this appears to be related to triangle CDE points.

3. Since ΔCDE is equilateral, all its angles are 60°. Angles adjacent to C due to ΔCDE minus the 98° in the rhombus require careful geometric rotation or consideration.

4. By a known geometric result for this figure, the exterior angle ∠BCD (98°) relates to angle x inside the triangle as:

x = `(180^circ - 98^circ)/2` = 41°.

This symmetry comes from the fact that two equal sides form equal angles opposite them in ΔCDE.

5. Angles y and z, being adjacent angles in the rhombus at vertex B and D formed by the construction of the equilateral triangle on side CD, are equal to x, so:

y = 41°

z = 41°

6. The small leftover angle x near vertex C inside the equilateral configuration corresponds to 11°, coming from the difference in angles formed by the equilateral triangle and the rhombus diagonal.

x = 11°  ...(Small angle caused by the difference around point C)

y = 41°  ...(Angle formed balancing the rhombus and equilateral triangle configurations)

z = 41°  ...(Equal to y by symmetry and rhombus properties)

In essence, the problem uses the property that the angle ∠BCD (98°) is split by the equilateral ΔCDE to yield these smaller angles and by considering the congruency and supplementary angle relationships, the values come out as x = 11°, y = 41°, z = 41°.