KLMN is an isosceles trapezium whose diagonals cut at X and KL is parallel to NM. If `angle KNL = 25^(@),angle KMN = 30^(@) angle KLN = 40^(@)`, find `angle NKX`

Answer

To solve the problem step by step, we will use the properties of isosceles trapeziums and the relationships between angles formed by parallel lines and transversals.

### Step 1: Understand the given information
We have an isosceles trapezium KLMN where KL is parallel to NM. The angles given are:
- \( \angle KNL = 25^\circ \)
- \( \angle KMN = 30^\circ \)
- \( \angle KLN = 40^\circ \)

### Step 2: Identify the angles at point X
Since KL is parallel to NM and KN is a transversal, we can use the property of alternate interior angles. Thus, we have:
- \( \angle KMN = \angle NKX = 30^\circ \)

### Step 3: Calculate the exterior angle at point K
In triangle KNL, we can find the exterior angle at K, which is given by the sum of the two opposite interior angles:
- \( \angle KNL + \angle KLN = 25^\circ + 40^\circ = 65^\circ \)
- Therefore, the exterior angle \( \angle KMN \) (which is also \( \angle NKX \)) is:
- \( \angle NKX + \angle KNL = 70^\circ \)

### Step 4: Set up the equation for triangle NKX
In triangle NKX, the sum of the angles is equal to \( 180^\circ \):
- \( \angle NKX + \angle KNL + \angle KXN = 180^\circ \)
- Substituting the known angles:
- \( \angle NKX + 25^\circ + 70^\circ = 180^\circ \)

### Step 5: Solve for angle NKX
Now, simplify the equation:
- \( \angle NKX + 95^\circ = 180^\circ \)
- Therefore, \( \angle NKX = 180^\circ - 95^\circ = 85^\circ \)

### Final Answer
Thus, the angle \( \angle NKX \) is \( 85^\circ \).

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