In the given triangle, CD is the bisector of ∠BCA. CD = DA. If ∠BDC = 76°, what is the degree measure of ∠CBD?
Answer & explanation
Correct answer: option 4
Concept used
The two angles of an isosceles triangle, opposite to equal sides are equal in measure.
Angle sum property = Sum of all the three angles of a triangle is \({180}^\circ\).
Calculations
In \(\Delta \) ABC
CD is the bisector of ∠BCA
⇒ ∠BCA = ∠DCA = \(\Theta \)
Since, CD = DA
⇒ ∠DCA = ∠CAD = \(\Theta \)
⇒ ∠BDC = \({76}^\circ\)
⇒ ∠BDC = ∠DCA + ∠CAD
⇒ \(\Theta \) + \(\Theta \) = \({76}^\circ\)
⇒ 2\(\Theta \) = \({76}^\circ\)
⇒ \(\Theta \) = \({38}^\circ\)
In \(\Delta \)CBD,
∠BCD + ∠CDB + ∠CBD = \({180}^\circ\)
⇒ \(\Theta \) + \({76}^\circ\) + ∠CDB = \({180}^\circ\)
⇒ \({38}^\circ\) + \({76}^\circ\) + ∠CDB = \({180}^\circ\)
⇒ ∠CDB = \({180}^\circ\) - \({114}^\circ\)
⇒ ∠CDB = \({66}^\circ\)
Therefore, answer is \({66}^\circ\)