suppose that $ \triangle c d e $ is isosceles with base $ \overline{e c} $.
suppose also that $ m \angle e=(2 x+35)^{circ} $ and $ m \angle c=(4 x+15)^{circ} $.
find the degree measure of each angle in the triangle.

Question

Accepted Answer

# Explanation:
## Step1: Identify equal angles in isosceles triangle
In isosceles triangle $ \triangle CDE $ with base $ EC $, the legs are $ DE $ and $ CD $, so the base angles are $ \angle E $ and $ \angle C $? Wait, no—wait, base is $ EC $, so the equal sides are $ DE $ and $ CD $, so the equal angles are $ \angle E $ and $ \angle C $? Wait, no, in an isosceles triangle with base $ EC $, the legs are $ DE $ and $ DC $, so the base angles are $ \angle E $ and $ \angle C $? Wait, the diagram shows $ DE $ and $ DC $ marked as equal (the tick marks), so $ DE = DC $, so the base is $ EC $, so the base angles are $ \angle E $ and $ \angle C $? Wait, no, in a triangle, the angles opposite equal sides are equal. So if $ DE = DC $, then the angles opposite them are $ \angle C $ and $ \angle E $, respectively. Wait, side $ DE $ is opposite $ \angle C $, and side $ DC $ is opposite $ \angle E $. So if $ DE = DC $, then $ \angle C = \angle E $? Wait, but the problem gives $ m\angle E = (2x + 35)^\circ $ and $ m\angle C = (4x + 15)^\circ $. Wait, maybe I made a mistake. Wait, the base is $ EC $, so the legs are $ DE $ and $ DC $, so the two equal sides are $ DE $ and $ DC $, so the angles opposite them are $ \angle C $ (opposite $ DE $) and $ \angle E $ (opposite $ DC $). Therefore, $ \angle C = \angle E $? But the given expressions are different. Wait, maybe the diagram has $ DE $ and $ DC $ as equal, so $ \angle E = \angle C $? Wait, but the problem says "isosceles with base $ EC $", so the legs are $ DE $ and $ DC $, so $ DE = DC $, so $ \angle E = \angle C $. Wait, but the problem gives $ m\angle E = (2x + 35) $ and $ m\angle C = (4x + 15) $. So set them equal: $ 2x + 35 = 4x + 15 $. Wait, solving: $ 35 - 15 = 4x - 2x $, $ 20 = 2x $, $ x = 10 $. Wait, but let's check. If $ x = 10 $, then $ m\angle E = 2(10) + 35 = 55^\circ $, $ m\angle C = 4(10) + 15 = 55^\circ $. Then the sum of angles in a triangle is $ 180^\circ $, so $ m\angle D + m\angle E + m\angle C = 180 $, so $ m\angle D = 180 - 55 - 55 = 70^\circ $. Wait, but let's confirm the isosceles triangle properties. Wait, maybe I mixed up the equal angles. Wait, the base is $ EC $, so the two equal sides are $ DE $ and $ DC $, so the base angles are $ \angle E $ and $ \angle C $, so they should be equal. So set $ 2x + 35 = 4x + 15 $.

## Step1: Set $ \angle E = \angle C $ (since $ DE = DC $)
$ 2x + 35 = 4x + 15 $

## Step2: Solve for $ x $
Subtract $ 2x $ from both sides: $ 35 = 2x + 15 $

Subtract 15 from both sides: $ 20 = 2x $

Divide by 2: $ x = 10 $

## Step3: Find $ m\angle E $ and $ m\angle C $
$ m\angle E = 2(10) + 35 = 55^\circ $

$ m\angle C = 4(10) + 15 = 55^\circ $

## Step4: Find $ m\angle D $
Sum of angles in triangle: $ m\angle D + m\angle E + m\angle C = 180^\circ $

$ m\angle D + 55 + 55 = 180 $

$ m\angle D + 110 = 180 $

$ m\angle D = 180 - 110 = 70^\circ $

# Answer:
$ m\angle C = 55^\circ $, $ m\angle E = 55^\circ $, $ m\angle D = 70^\circ $

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QUESTION IMAGE

Question

Explanation:

Step1: Identify equal angles in isosceles triangle

Step1: Set \( \angle E = \angle C \) (since \( DE = DC \))

Step2: Solve for \( x \)

Step3: Find \( m\angle E \) and \( m\angle C \)

Step4: Find \( m\angle D \)

Answer: