Question

in (\triangle nop), (overline{np}) is extended through point (p) to point (q), (mangle opq=(7x - 11)^{circ}), (mangle nop=(2x + 14)^{circ}), and (mangle pno=(3x + 7)^{circ}). find (mangle opq).

Explanation:

Step1: Use exterior angle theorem

The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle NOP\), \(\angle OPQ\) is an exterior angle. So, \(m\angle OPQ=m\angle NOP + m\angle PNO\).
Substitute the given expressions: \((7x - 11)=(2x + 14)+(3x + 7)\).

Step2: Solve the equation for \(x\)

Simplify the right - hand side: \((2x + 14)+(3x + 7)=2x+3x + 14 + 7=5x+21\).
The equation becomes \(7x-11 = 5x + 21\).
Subtract \(5x\) from both sides: \(7x-5x-11=5x - 5x+21\), \(2x-11 = 21\).
Add \(11\) to both sides: \(2x-11 + 11=21 + 11\), \(2x=32\).
Divide both sides by \(2\): \(x=\frac{32}{2}=16\).

Step3: Find \(m\angle OPQ\)

Substitute \(x = 16\) into the expression for \(m\angle OPQ\).
\(m\angle OPQ=(7x - 11)^{\circ}\).
\(m\angle OPQ=7\times16-11\).
\(m\angle OPQ=112 - 11\).
\(m\angle OPQ = 101^{\circ}\).

Answer:

\(101^{\circ}\)