Question

In ∆ABC the angle bisectors drawn from vertices A and B intersect at point D. Find m∠ADB if: m∠C=130°

Answer 1

Answer:

∠ADB=155°.

Step-by-step explanation:

In Δ ABC, let A = x°

By angle-sum property.

A+B+C=180°

But, it is given that C=130°

So, x+B+130=180

B=180-130-x

B=50-x

Since AD and BD are internal bisectors of A and B,

∠DAB=x/2 and

∠DBA=

In ΔADB, by angle-sum property,

∠DBA+∠DAB+∠ADB=180°

+∠ADB=180°

25+∠ADB=180°

∠ADB=180-25=155°

Hence, ∠ADB=155°.

Answer 2

Answer:

  m∠ADB = 155°

Step-by-step explanation:

  m∠ADB = (m∠C)/2 + 90° = 130°/2 + 90°

  m∠ADB = 155°

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Reference brainly.com/question/17443119 for the derivation.