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Question
describing the exterior a of a triangle continued
(17x + 2)°
(9x + 10)°
(15x + 20)°
m∠bef =
m∠abc =
Step1: Use the exterior angle sum formula
The sum of the exterior angles of a triangle is \(360^{\circ}\). So, \((17x + 2)+(9x + 10)+(15x + 20)=360\).
Step2: Simplify the equation
Combine like - terms: \(17x+9x + 15x+2 + 10+20 = 360\), which gives \(41x+32 = 360\).
Step3: Solve for \(x\)
Subtract 32 from both sides: \(41x=360 - 32=328\). Then \(x=\frac{328}{41}=8\).
Step4: Find \(m\angle BEF\)
Substitute \(x = 8\) into \(17x + 2\). So, \(m\angle BEF=17\times8+2=136 + 2=138^{\circ}\).
Step5: Find \(m\angle ABC\)
\(\angle ABC\) and \((15x + 20)\) are supplementary (linear - pair). Substitute \(x = 8\) into \(15x + 20\), we get \(15\times8+20=120 + 20 = 140\). Then \(m\angle ABC=180-(15x + 20)=180-(15\times8 + 20)=180 - 140=40^{\circ}\).
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\(m\angle BEF = 138^{\circ}\), \(m\angle ABC=40^{\circ}\)