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# help

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What is the angle colored purple in degrees? Dec 26, 2019

#2
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If it is 2D....let's label the figure like this: △ABE  is an equilateral triangle, so every interior angle is 60°, and so

m∠EAB  =  60°

m∠ABE  =  60°

We are given that

m∠EBC  =  90°

Now we can determine that

m∠ABC  =  60° + 90°

m∠ABC  =  150°

△ABC is an isosceles triangle. so base angles are congruent, and so

m∠BCA  =  m∠BAC

The sum of the measures of the interior angles in a triangle is 180°, so

m∠BAC  +  m∠BCA  +  m∠ABC  =  180°

Substitute  m∠BAC  in for  m∠BCA  and  150°  in for  m∠ABC

m∠BAC  +  m∠BAC  +     150°    =   180°

Combine like terms

2(m∠BAC)  +  150°  =  180°

Subtract  150°  from both sides of the equation.

2(m∠BAC)  =  30°

Divide both sides of the equation by  2

m∠BAC  =  15°

In the same way, we can determine that

Then

m∠DAC   =   m∠EAB  - m∠EAD - m∠BAC   =   60° - 15° - 15°   =   30°

Dec 26, 2019

#1
0

Is the a 2d object or is it a representation of a 3D object.

What object is it?

Dec 26, 2019
#2
+2

If it is 2D....let's label the figure like this: △ABE  is an equilateral triangle, so every interior angle is 60°, and so

m∠EAB  =  60°

m∠ABE  =  60°

We are given that

m∠EBC  =  90°

Now we can determine that

m∠ABC  =  60° + 90°

m∠ABC  =  150°

△ABC is an isosceles triangle. so base angles are congruent, and so

m∠BCA  =  m∠BAC

The sum of the measures of the interior angles in a triangle is 180°, so

m∠BAC  +  m∠BCA  +  m∠ABC  =  180°

Substitute  m∠BAC  in for  m∠BCA  and  150°  in for  m∠ABC

m∠BAC  +  m∠BAC  +     150°    =   180°

Combine like terms

2(m∠BAC)  +  150°  =  180°

Subtract  150°  from both sides of the equation.

2(m∠BAC)  =  30°

Divide both sides of the equation by  2

m∠BAC  =  15°

In the same way, we can determine that

Then

m∠DAC   =   m∠EAB  - m∠EAD - m∠BAC   =   60° - 15° - 15°   =   30°

hectictar Dec 26, 2019
#3
+1

Triangle side  >  s = 2

---II---   angle  >  q = 60°

---II---  height  >  h = ?                   h = tan(60°)  = 1.732

Square side    >  a = 2

Total height     >  H = ?                  H = a+h      H = 2+1.732     H = 3.732

Desired angle >  A = ?                   tan(q/2) = (a/2)/H  = 15        A = 15*2   = 30° Dec 27, 2019