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Question
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in the figure, there is a triangle cba with angle at c being 38 degrees. there is a straight line ebaf, with e---b---a---f on the line. angle at b inside the triangle is (4x + 2) degrees, and angle at a outside the triangle (on the line af side) is (7x + 13) degrees.
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 \(ABC\), \(\angle CAF=(7x + 13)^{\circ}\) is an exterior angle, and the two non - adjacent interior angles are \(\angle C = 38^{\circ}\) and \(\angle ABC=(4x + 2)^{\circ}\). So we can set up the equation:
\(7x+13=38+(4x + 2)\)
Step2: Simplify the equation
First, simplify the right - hand side of the equation: \(38+(4x + 2)=4x+40\). So our equation becomes \(7x + 13=4x+40\).
Step3: Solve for \(x\)
Subtract \(4x\) from both sides: \(7x-4x+13=4x - 4x+40\), which simplifies to \(3x+13 = 40\).
Then subtract 13 from both sides: \(3x+13-13=40 - 13\), so \(3x=27\).
Divide both sides by 3: \(x=\frac{27}{3}=9\).
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\(x = 9\)