Question

what is the value of c?
11c+13°
13c-50°
7c-40°
write your answer as an integer or as a decimal rounded to the nearest tenth.
c = □°

Explanation:

Step1: Identify the theorem

We use the Exterior Angle Theorem, which states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So, \(11c + 13=(13c-50)+(7c - 40)\).

Step2: Simplify the right - hand side

First, combine like terms on the right - hand side: \((13c-50)+(7c - 40)=13c+7c-50 - 40=20c-90\). So our equation becomes \(11c + 13=20c-90\).

Step3: Solve for c

Subtract \(11c\) from both sides: \(13=20c-11c - 90\), which simplifies to \(13 = 9c-90\). Then add 90 to both sides: \(13 + 90=9c\), so \(103 = 9c\). Now divide both sides by 9: \(c=\frac{103}{9}\approx11.4\) (rounded to the nearest tenth) or if we solve exactly, \(c = \frac{103}{9}\approx11.4\). Wait, let's check the calculation again. Wait, \(11c + 13=(13c - 50)+(7c-40)\)
\(11c+13 = 13c+7c-50 - 40\)
\(11c + 13=20c-90\)
Subtract \(11c\) from both sides: \(13=9c - 90\)
Add 90 to both sides: \(13 + 90=9c\)
\(103 = 9c\)
\(c=\frac{103}{9}\approx11.4\)? Wait, no, wait \(13c+7c = 20c\), \(-50-40=-90\). Then moving terms: \(11c+13 = 20c-90\)
\(13 + 90=20c-11c\)
\(103 = 9c\)
\(c=\frac{103}{9}\approx11.4\)? Wait, but let's check with the angle measures. Wait, maybe I made a mistake in the theorem application. Wait, the exterior angle is equal to the sum of the two remote interior angles. So the exterior angle \(11c + 13\) should equal the sum of the two non - adjacent interior angles \(13c-50\) and \(7c - 40\). Wait, but let's check the arithmetic again.
Wait, \(13c+7c=20c\), \(-50-40 = - 90\). So equation: \(11c + 13=20c-90\)
Subtract \(11c\): \(13=9c - 90\)
Add 90: \(103 = 9c\)
\(c=\frac{103}{9}\approx11.4\). Wait, but let's check if the angles are positive.
For \(13c-50\): \(13\times11.4-50=148.2 - 50 = 98.2>0\)
For \(7c-40\): \(7\times11.4-40 = 79.8-40=39.8>0\)
For \(11c + 13\): \(11\times11.4+13=125.4 + 13=138.4\)
And \(98.2+39.8 = 138\), which is close (due to rounding). Wait, maybe I made a mistake in the equation setup. Wait, no, the exterior angle theorem is correct. Wait, \(13c-50+7c - 40=20c-90\), and the exterior angle is \(11c + 13\). So \(11c+13=20c - 90\)
\(13 + 90=20c-11c\)
\(103 = 9c\)
\(c=\frac{103}{9}\approx11.4\). Wait, but let's do the division: \(103\div9 = 11.444\cdots\), so rounded to the nearest tenth is \(11.4\). But wait, maybe I messed up the signs. Wait, let's re - derive the equation.
The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So if the exterior angle is \(11c + 13\), and the two non - adjacent interior angles are \(13c-50\) and \(7c - 40\), then:
\(11c+13=(13c - 50)+(7c-40)\)
\(11c+13=13c + 7c-50 - 40\)
\(11c+13=20c-90\)
Subtract \(11c\) from both sides:
\(13 = 9c-90\)
Add 90 to both sides:
\(13 + 90=9c\)
\(103 = 9c\)
\(c=\frac{103}{9}\approx11.4\) (rounded to the nearest tenth)

Answer:

\(11.4\)