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In /_\PQR, bar(PR) is extended through point R to point S,m/_PQR=(x+12)^(@), m/_RPQ=(2x+7)^(@), and m/_QRS=(7x-17)^(@). Find m/_PQR. Answer:

In △PQR,PR‾ \triangle \mathrm{PQR}, \overline{P R} is extended through point R \mathrm{R} to point S,m∠PQR=(x+12)∘ \mathrm{S}, \mathrm{m} \angle P Q R=(x+12)^{\circ} , m∠RPQ=(2x+7)∘ \mathrm{m} \angle R P Q=(2 x+7)^{\circ} , and m∠QRS=(7x−17)∘ \mathrm{m} \angle Q R S=(7 x-17)^{\circ} . Find m∠PQR \mathrm{m} \angle P Q R .\newlineAnswer:

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Q. In △PQR,PR‾ \triangle \mathrm{PQR}, \overline{P R} is extended through point R \mathrm{R} to point S,m∠PQR=(x+12)∘ \mathrm{S}, \mathrm{m} \angle P Q R=(x+12)^{\circ} , m∠RPQ=(2x+7)∘ \mathrm{m} \angle R P Q=(2 x+7)^{\circ} , and m∠QRS=(7x−17)∘ \mathrm{m} \angle Q R S=(7 x-17)^{\circ} . Find m∠PQR \mathrm{m} \angle P Q R .\newlineAnswer:
  1. Understand angles relationship: Understand the relationship between the angles in the problem.\newlineIn any triangle, the sum of the interior angles is always 180180 degrees. Additionally, the exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.
  2. Set up equation: Set up the equation using the exterior angle theorem.\newlineThe measure of angle QRS (exterior angle) is equal to the sum of the measures of angles PQR and RPQ.\newlineSo, we have the equation: m∠QRS=m∠PQR+m∠RPQm\angle QRS = m\angle PQR + m\angle RPQ\newlineSubstitute the given expressions: (7x−17)∘=(x+12)∘+(2x+7)∘(7x - 17)^\circ = (x + 12)^\circ + (2x + 7)^\circ
  3. Combine terms and solve: Combine like terms and solve for xx.
    (7x−17)=(x+12)+(2x+7)(7x - 17) = (x + 12) + (2x + 7)
    7x−17=3x+197x - 17 = 3x + 19
    7x−3x=19+177x - 3x = 19 + 17
    4x=364x = 36
    x=9x = 9
  4. Substitute value for measure: Substitute the value of xx back into the expression for m∠PQRm\angle PQR to find its measure.\newlinem∠PQR=(x+12)∘m\angle PQR = (x + 12)^\circ\newlinem∠PQR=(9+12)∘m\angle PQR = (9 + 12)^\circ\newlinem∠PQR=21∘m\angle PQR = 21^\circ

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