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 length, this bad boy is an acute scalene triangle. |