Math, asked by hkishor60gmailcom, 1 month ago

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Answered by bharathidoure
2

In △ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.Read more on Sarthaks.com - https://www.sarthaks.com/687772/in-abc-a-50and-bc-is-produced-to-a-point-d-the-bisectors-of-abcand-acd-meet-at-e-find-e

Step-by-step explanation:

∠ACD = ∠A + ∠B. (Exterior angle is equal to the sum of two opposite interior angles.) We know that the sum of all three angles of a triangle is 180°. Therefore, for the given triangle, we know that the sum of the angles = 180° ∠ABC + ∠BCA + ∠CAB = 180° ∠A + ∠B + ∠BCA = 180° ∠BCA = 180° – (∠A + ∠B) But we know that EC bisects ∠ACD Therefore ∠ECA = ∠ACD/2 ∠ECA = (∠A + ∠B)/2 [∠ACD = (∠A + ∠B)] But EB bisects ∠ABC ∠EBC = ∠ABC/2 = ∠B/2 ∠EBC = ∠ECA + ∠BCA ∠EBC = (∠A + ∠B)/2 + 180° – (∠A + ∠B) If we use same steps for △EBC, then we get, ∠B/2 + (∠A + ∠B)/2 + 180° – (∠A + ∠B) + ∠BEC = 180° ∠BEC = ∠A + ∠B – (∠A + ∠B)/2 – ∠B/2 ∠BEC = ∠A/2 ∠BEC = 50°/2 = 25°Read more on Sarthaks.com - https://www.sarthaks.com/687772/in-abc-a-50and-bc-is-produced-to-a-point-d-the-bisectors-of-abcand-acd-meet-at-e-find-e

Answered by shrutishreya2007
0

Answer:

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