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Find the angles look at the picture

Find the angles look at the picture-example-1

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Answer:

  • m∠B = 42°
  • m∠E = 69°
  • m∠ECD = 42°
  • m∠ACD = 138°

Explanation:

Given;

  • ∆ ABC is an isoceles triangle.
  • ∆CDE is an isoceles triangle.
  • m∠A = 96°

To find:

  • m∠B = ?
  • m∠E = ?
  • m∠ECD = ?
  • m∠ACD = ?

Solution:

Since the base angle of isoceles triangle are equal.

So,

m∠B = m∠C

We know that

Sum of interior angles of a triangle is equal to 180°.

So,

m∠A + m∠B + m∠C = 180°

Substitute known value

96° + m∠C + m∠C = 180°

96° + 2m∠C = 180°

2m∠C = 180° - 96°

2m∠C = 84°


\sf m \angle C =(84^\circ)/(2)

m∠C = 42°

Since m∠B = m∠C so,

m∠B = 42°

Similarly

Sum of interior angles of a triangle is equal to 180°.

so,

m∠CDE + m∠CED + m∠ECD = 180°

Here

∆ CDE is an isoceles triangle.

So, it's base angle are equal, which is:

m∠CDE = m∠CED

And

m∠C = m∠ECD = 42° Vertically Opposite Angle are equal.

Substitute these known value:

Here

m∠CED + m∠CED + 42° = 180°

2m∠CED + 42° = 180°

2m∠CED = 180° - 42°

2m∠CED = 138°


\sf m\angle CED =(138^\circ)/(2)

m∠CED = 69°

So,

m∠CDE = 69°

And finally

m∠ACD and m∠C are linear pair and they are supplementary.

So,

m∠ACD + m∠C = 180°

m∠ACD + 42° = 180°

m∠ACD = 180° - 42°

m∠ACD = 138°

Therefore, final answer is:

  • m∠B = 42°
  • m∠E = 69°
  • m∠ECD = 42°
  • m∠ACD = 138°
User Prateek Prasad
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