Question

A 80° C B \angle A + \angle B + \angle C = 180\degree (i) AB = BC \angle B = \angle C (ii) From equation (i) and (ii) \angle A + \angle B + \angle C = 180\degree 80\degree + 2\angle B = 180\degree 2\angle B = 180\degree - 80\degree = 100\degree \angle B = \frac{100\degree}{2} \angle B = 50\degree

          A
80°
C
B
\angle A + \angle B + \angle C = 180\degree (i)
AB = BC
\angle B = \angle C (ii)
From equation (i) and (ii)
\angle A + \angle B + \angle C = 180\degree
80\degree + 2\angle B = 180\degree
2\angle B = 180\degree - 80\degree
= 100\degree
\angle B = \frac{100\degree}{2}
\angle B = 50\degree

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A
80°
C
B
∠A + ∠B + ∠C = 180(i)
AB = BC
∠B = ∠C (ii)
From equation (i) and (ii)
∠A + ∠B + ∠C = 18080+ 2∠B = 1802∠B = 180- 80= 100∠B = (100)/(2)
∠B = 50

Added by Joseph M.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

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Solved by Expert Mohammed Nadhir

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In triangle ABC, we are given that angle A is 80°. Show more…

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