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Given: ΔABC
m∠1=m∠2
D∈ AC , BD = DC
m∠BDC=100°
Find: m∠A, m∠B, m∠C

Given: ΔABC m∠1=m∠2 D∈ AC , BD = DC m∠BDC=100° Find: m∠A, m∠B, m∠C-example-1

2 Answers

3 votes

Answer:

m<A=60 degrees, m<B=80 degrees, m<C=60 degrees.

Explanation:

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User Rolan
by
5.5k points
7 votes

Answer:

  • m∠A = 60°, m∠B = 80°, m∠C = 40°

Explanation:

Since BD = DC, the triangle BDC is isosceles, so

  • m∠C = m∠2

Having m∠BDC = 100°, we can find m∠C

  • m∠C = 1/2(180° - 100°) = 1/2(80°) = 40°

Finding angle B

  • m∠B = m∠1 + m∠2 = 2*m∠2= 2*40° = 80°

Finding angle A

  • m∠BDA = 180° - m∠BDC = 180° - 100° = 80°
  • m∠A = 180° - (m∠1 + m∠BDA) = 180° - (40° + 80°) = 180° - 120° = 60°
User David Rissato Cruz
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4.7k points