Math  /  Geometry

QuestionIf EC\overline{EC} is an altitude of AED\triangle AED, and AC=CDAC = CD, with m1=2x+7m \angle 1 = 2x + 7 and m2=3x+13m \angle 2 = 3x + 13, find m1m \angle 1 and m2m \angle 2.

Studdy Solution

STEP 1

1. EC \overline{EC} is an altitude of AED \triangle AED , meaning it is perpendicular to AD \overline{AD} .
2. Since AC=CD AC = CD , ACD \triangle ACD is isosceles.
3. Angles 1 \angle 1 and 2 \angle 2 are complementary because they are formed by the intersection of the altitude with the triangle.

STEP 2

1. Use the property of complementary angles.
2. Set up an equation to solve for x x .
3. Calculate m1 m \angle 1 and m2 m \angle 2 .

STEP 3

Since EC \overline{EC} is an altitude, it forms right angles with AD \overline{AD} . Therefore, 1+2=90 \angle 1 + \angle 2 = 90^\circ .

STEP 4

Set up the equation using the expressions given for m1 m \angle 1 and m2 m \angle 2 :
(2x+7)+(3x+13)=90 (2x + 7) + (3x + 13) = 90

STEP 5

Combine like terms:
2x+3x+7+13=90 2x + 3x + 7 + 13 = 90 5x+20=90 5x + 20 = 90

STEP 6

Solve for x x :
5x+20=90 5x + 20 = 90 5x=9020 5x = 90 - 20 5x=70 5x = 70 x=705 x = \frac{70}{5} x=14 x = 14

STEP 7

Substitute x=14 x = 14 back into the expressions for m1 m \angle 1 and m2 m \angle 2 :
m1=2x+7=2(14)+7=28+7=35 m \angle 1 = 2x + 7 = 2(14) + 7 = 28 + 7 = 35^\circ

STEP 8

Calculate m2 m \angle 2 :
m2=3x+13=3(14)+13=42+13=55 m \angle 2 = 3x + 13 = 3(14) + 13 = 42 + 13 = 55^\circ
The measures of the angles are:
m1=35 m \angle 1 = 35^\circ m2=55 m \angle 2 = 55^\circ

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