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Question
unit 1 exam on lessons 1.1 to 1.12 (33 points possible)
- (5 points) in the diagram below of △hqp, side $overline{hp}$ is extended through p to t, $mangle qpt=6x + 20$, $mangle hqp=x + 40$, and $mangle phq=4x - 5$. find the value of x.
Step1: Apply exterior - angle theorem
The exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle HQP\), \(\angle QPT\) is an exterior angle, so \(m\angle QPT=m\angle HQP + m\angle PHQ\).
\[6x + 20=(x + 40)+(4x-5)\]
Step2: Simplify the right - hand side
Combine like terms on the right - hand side: \((x + 40)+(4x-5)=x+4x + 40-5=5x + 35\). So the equation becomes \(6x + 20=5x + 35\).
Step3: Solve for \(x\)
Subtract \(5x\) from both sides of the equation: \(6x-5x+20=5x-5x + 35\), which simplifies to \(x+20 = 35\). Then subtract 20 from both sides: \(x=35 - 20\).
\[x = 15\]
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\(15\)