Question

the figure below is a rhombus. find the value of x. 102° x° x = ?°

Explanation:

Step1: Recall properties of a rhombus

In a rhombus, adjacent angles are supplementary, and the diagonal bisects the angles. Also, the sum of adjacent angles in a rhombus is \(180^\circ\). First, find the adjacent angle to \(102^\circ\). Let the adjacent angle be \(y\), then \(102^\circ + y= 180^\circ\), so \(y = 180^\circ - 102^\circ=78^\circ\)? Wait, no, actually, the diagonal of a rhombus bisects the angles. Wait, the angle given is \(102^\circ\), and the diagonal splits the angle? Wait, no, in a rhombus, the diagonal bisects the vertex angles. Wait, the sides are equal, and the diagonal divides the rhombus into two isosceles triangles. Also, consecutive angles in a rhombus are supplementary. Wait, the angle at the top left is \(102^\circ\), so the angle adjacent to it (top right and bottom left) would be supplementary? Wait, no, in a rhombus, opposite angles are equal, and consecutive angles are supplementary. So the angle adjacent to \(102^\circ\) (let's say the angle at the top right) is \(180 - 102=78^\circ\)? Wait, no, the diagonal is drawn from the top right vertex to the bottom left vertex. Wait, the triangle formed by the diagonal: in a rhombus, all sides are equal, so the triangle is isosceles. Also, the diagonal bisects the angles. Wait, the angle at the top left is \(102^\circ\), so the diagonal will bisect the angle? No, wait, the diagonal that connects the two vertices with the angle \(102^\circ\) and the other angle. Wait, maybe a better approach: in a rhombus, the diagonal bisects the internal angles. Wait, the angle given is \(102^\circ\), and the diagonal is splitting the angle? Wait, no, the angle at the top left is \(102^\circ\), and the diagonal goes from the top right to the bottom left. So the triangle on the top: the two sides of the rhombus are equal, so the triangle is isosceles. Also, the sum of angles in a triangle is \(180^\circ\). Wait, another property: in a rhombus, adjacent angles are supplementary, so the angle next to \(102^\circ\) (let's say the angle at the top right) is \(180 - 102 = 78^\circ\)? No, wait, no, the angle at the top left is \(102^\circ\), so the angle at the bottom left is also \(102^\circ\) (opposite angles), and the angles at the top right and bottom right are equal, each \(180 - 102=78^\circ\). Now, the diagonal from the top right to the bottom left: this diagonal will bisect the angles at the top right and bottom left. Wait, the angle at the top right is \(78^\circ\), so the diagonal bisects it into two angles of \(x^\circ\). Wait, no, the angle marked \(x^\circ\) is at the top right vertex, between the top side and the diagonal. So the diagonal bisects the angle at the top right. Since the angle at the top right is \(180 - 102 = 78^\circ\)? Wait, no, that can't be. Wait, let's correct: in a rhombus, consecutive angles are supplementary. So angle \(A +\) angle \(B = 180^\circ\). If angle \(A = 102^\circ\), then angle \(B = 78^\circ\). The diagonal of the rhombus bisects angle \(B\) (and angle \(D\), which is equal to angle \(B\)). So the diagonal splits angle \(B\) into two equal angles, each of measure \(x\). So \(x=\frac{180 - 102}{2}\)? Wait, \(180 - 102 = 78\), then \(78\div2 = 39\). Wait, that makes sense. Because the diagonal bisects the angle, and the consecutive angles are supplementary. So the angle adjacent to \(102^\circ\) is \(78^\circ\), and the diagonal bisects it, so \(x = \frac{78}{2}=39\).

Step2: Calculate \(x\)

Consecutive angles in a rhombus are supplementary, so the angle adjacent to \(102^\circ\) is \(180^\circ - 102^\circ = 78^\circ\). The diagonal of a rhombus bisects this angle, so \(x=\frac{78^\circ}{2}=39^\circ\).

Answer:

\(39\)