Question

In the diagram below of ΔACD, B is a point on AC⎯⎯⎯⎯⎯such that ΔADB is an equilateral triangle, and ΔDBC is an isosceles triangle with DB⎯⎯⎯⎯⎯≅BC⎯⎯⎯⎯⎯.

Find m∠C. Explain your work/thought process.

Answer 1

Answer:

equilateral triangle all angles add up to 180 and are equal so each angle is 60 degrees

angles on a line add up to 180.

angle CBD = 180-60 = 120

DBC is an isosceles triangle so it's base angles are the same.

180-120= 60

60/2=30

Answer 2

The isosceles triangle ΔDBC have equal base angles.

The measure of angle C is 30 degrees

ΔADB is an equilateral triangle.

So, we have:

$\mathbf{\angle ADC = \angle BAD = \angle DBA = 60}$

Next, we calculate $\mathbf{\angle DBC}$

$\mathbf{\angle DBC =180 - \angle DBA}$ -- angle on a straight line

So, we have:

$\mathbf{\angle DBC =180 - 60}$

$\mathbf{\angle DBC =120}$

ΔDBC is an isosceles triangle.

So, we have:

$\mathbf{\angle C = \angle D}$

This is then calculated as:

$\mathbf{\angle C + \angle C = 180 - 120}$

$\mathbf{2\angle C = 60}$

Divide both sides by 2

$\mathbf{\angle C = 30}$

Hence, the measure of angle C is 30 degrees

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