Rotating Triangles

Geometry Level 1

Let $\triangle ABC$ and $\triangle CDE$ be equilateral triangles of the same size, and $\angle BCD=80^\circ$ between them. Find the measure of $\angle BAD$ in degrees.

The answer is 40.

3 solutions

Paola Ramírez
May 17, 2016

Relevant wiki: Properties of Isosceles Triangles

$\angle ACD=140°$ and $\triangle ACD$ is isosceles $\Rightarrow \angle DAC=\angle CDA=20°$

$\angle BAD=60°-20°=\boxed{40°}$

How is AC=CD or how is the triangle isosceles because it is not given that they are equal or congruent equilateral triangle?

Satwik Murarka - 5 years ago

It needs to be specified in the problem statement that the triangle are the same size.

A Former Brilliant Member - 5 years ago

Thanks. I have edited the problem accordingly.

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Calvin Lin Staff - 5 years ago

A Former Brilliant Member
Oct 29, 2017

$\alpha = 180-60-80=$ $\boxed{40}$

This is a very long solution. You computed many angles.

A Former Brilliant Member - 3 years, 7 months ago

A Former Brilliant Member
Dec 16, 2017

Consider isosceles $\triangle ACD$ .

$\angle ACD = 60 + 80 = 140^\circ$

Since $AC=CD$ , $\angle ADC= \angle DAC=\dfrac{180-140}{2}=20^\circ$ . It follows that $\boxed{\angle \alpha = 40^\circ}$ .

Rotating Triangles

The answer is 40.

3 solutions

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