Question

2. In the adjoining figure, ZB = 70° and ZA = 50°. If the bisector of ZC meets AB in D, then find ZADC.​

Answer 1

Answer:

onsider the ΔABC, We know that, sum of angles of triangle is equal to 180o. So, ∠Α + ∠B + ∠C = 180o 50o + 70o + ∠C = 180o ∠C + 120o = 180o ∠C = 180o – 120o ∠C = 60o Since CD bisects ∠C, So, ∠DCB = ∠ΑCD = ½ ∠C = 60o/2 = 30o Now, consider ΔBDC From exterior angle property, ∠ACD = ∠DBC + ∠DCB ∠ACD = 70o + 30o = 100oRead more on Sarthaks.com - https://www.sarthaks.com/792476/in-abc-50-b-70-and-bisector-of-c-meets-ab-in-d-fig-measure-of-adc-is