In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then,what is the measure of ∠B ?
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In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then,what is the measure of ∠B ?
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In ∆ABC, ∠Α = 100°, AD bisects ∠A and AD⊥BC. Then, ∠B is equal to (a) 80° (b) 20° (c) 40° (d) 30°
(c) 40o Consider the triangle ABC, From the figure, AD bisects ∠Α Then, ∠BAD = 50o ∠DAC = 50o So, AD⊥BC ∠ΑDC = 90o Consider the ∆ABD, ...
Step-by-step explanation:
Consider the triangle ABC, From the figure, AD bisects ∠Α Then, ∠BAD = 50o ∠DAC = 50o So, AD⊥BC ∠ΑDC = 90o Consider the ∆ABD, From the rule of exterior angle property of triangle, ∠ΑDC = ∠ΑBD + ∠BAD 90o = ∠ΑBD + 50o ∠ΑBD = 90o – 50o = 40 degrees please mark as brainlliest