Since we know that ∠*ABC* is an exterior angle of ∆*BCD*, we can use the Exterior Angle Theorem to help us find ∠*BCD*. Remember the Exterior Angle Theorem? m∠*ABC* = m∠*BCD* + m∠*CDB* Well, we know that m∠*ABC* = 120, but what about ∠*CDB*? Luckily, its exterior angle, ∠*CDF*, is the same as ∠*CDE* + ∠*EDF*. m∠*CDB* = 180 – (m∠*CDE* + m∠*EDF*) Instead of substituting equations within equations (equation-ception?), we can solve them before we plug them back into the equation we had before. We already know that the angles in an equilateral triangle are 60° in measure, so that takes care of m∠*EDF*. That means it's time for some number action. m∠*CDB* = 180 – (64 + 60) m∠*CDB* = 56 Now we can use that to find m∠*BCD*. m∠*ABC* = m∠*BCD* + m∠*CDB* 120 = m∠*BCD* + 56 m∠*BCD* = 64 The remaining angle, ∠*CBD*, can be calculated using the Angle Sum Theorem or its 120° supplementary angle. Either way, we should get 60°. That means our triangle has angles of 56°, 60°, and 64°. Three acute angles make an acute triangle, and that's what we have here. Also, since each of the sides has a different |