QUESTION IMAGE
Question
what is the value of x?
Step1: Use the property of isosceles triangle
Since two sides are equal (both \(s - 50^{\circ}\)), the base - angles of the isosceles triangle are equal.
Step2: Use the exterior - angle property
The exterior angle \(x\) of a triangle is equal to the sum of the two non - adjacent interior angles. Here, the two non - adjacent interior angles are both \(s - 50^{\circ}\). So \(x=(s - 50^{\circ})+(s - 50^{\circ})\). But also, for the straight - line adjacent to \(x\) and \(s\), \(s+x = 180^{\circ}\). Since the triangle is isosceles, the two non - vertex angles are equal. Let's use another approach. The sum of angles in a triangle: if the vertex angle is \(y\), and the two equal angles are \(a=s - 50^{\circ}\). Then \(y+2a=180^{\circ}\). And \(x = 180^{\circ}-y\). Substituting \(y = 180^{\circ}-2a\) into \(x = 180^{\circ}-y\), we get \(x=2a\). Since \(a=s - 50^{\circ}\), and also, if we assume the triangle has two sides of length \(s\) (the notation is a bit unclear, but based on the equal side - angle markings). A more straightforward way: the exterior angle \(x\) of an isosceles triangle (where the two non - exterior adjacent angles are equal) is equal to \(100^{\circ}\) (because if we assume the two equal angles in the triangle are \(z\), and using the exterior - angle property \(x = 2z\), and if we consider the sum of angles in a triangle: if we assume the problem is based on a standard isosceles - triangle exterior - angle setup. Let's use the exterior - angle formula for a triangle. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. In an isosceles triangle (two sides equal, so two angles equal). Let the two equal angles be \(A\) and \(A\). Then \(x=A + A\). If we assume the problem is set up such that \(A = 50^{\circ}\) (a mis - notation in the problem, likely the angle value is \(50^{\circ}\) instead of \(s - 50^{\circ}\) due to a drawing error). Using the exterior - angle property of a triangle \(x=2\times50^{\circ}\)
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