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Question
what kind of triangle is △bcd? explain. c 70° 1 2 b d 130° a. right triangle; m∠2 = 90 b. acute triangle; m∠cdb = 180 - 130 = 50 and m∠2 = 180 - 70 - 50 = 60 c. obtuse triangle; m∠cdb = 180 - 70 = 110 d. isosceles triangle; m∠cdb = m∠2
Step1: Find angle $\angle CDB$
Since an exterior - angle and its adjacent interior - angle are supplementary, if the exterior angle at $D$ is $130^{\circ}$, then $m\angle CDB=180 - 130=50^{\circ}$.
Step2: Find angle $\angle 2$
Using the angle - sum property of a triangle ($m\angle C + m\angle CDB+m\angle 2 = 180^{\circ}$), and given $m\angle C = 70^{\circ}$ and $m\angle CDB = 50^{\circ}$, we have $m\angle 2=180-(70 + 50)=60^{\circ}$.
Step3: Classify the triangle
Since $m\angle C = 70^{\circ}$, $m\angle CDB = 50^{\circ}$, and $m\angle 2 = 60^{\circ}$, all angles of $\triangle BCD$ are less than $90^{\circ}$. So, it is an acute triangle.
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B. Acute triangle; $m\angle CDB = 180 - 130 = 50$ and $m\angle 2 = 180 - 70 - 50 = 60$