Question

b - aaa02
asures of polygons
the polygon interior angle sum theorem
what is the measure of \\( \angle b \\) of the pentagon?

Explanation:

Step1: Calculate the sum of interior angles

The formula for the sum of interior angles of an \(n -\)sided polygon is \((n - 2)\times180^{\circ}\). For a pentagon (\(n=5\)), the sum \(S=(5 - 2)\times180^{\circ}=540^{\circ}\).

Step2: Set up an equation to find \(\angle B\)

Let \(\angle B=x\). We know that \(\angle A = 92^{\circ}\), \(\angle C=89^{\circ}\), \(\angle D = 98^{\circ}\), \(\angle E=150^{\circ}\). Then \(x+92^{\circ}+89^{\circ}+98^{\circ}+150^{\circ}=540^{\circ}\).

Step3: Solve the equation

Simplify the left - hand side: \(x+(92 + 89+98 + 150)^{\circ}=x + 429^{\circ}\).
So, \(x+429^{\circ}=540^{\circ}\). Subtract \(429^{\circ}\) from both sides: \(x=540^{\circ}-429^{\circ}=111^{\circ}\).

Answer:

\(111\)