in triangle abc , angle b = 45 degree angle c = 55 degree and bisector af angle a meet bc at point d find the measure of angle adb and angle adc

plx help

Given: In ΔABC

∠B = 45°, ∠C = 55°

Now,

∠A + ∠B + ∠C = 180°  (Angle sum property)

⇒ ∠A + 45° + 55° = 180°

⇒ ∠A = 180° – 100° = 80°

Also, AD is bisector of ∠A.

 

Now,

∠BAD + ∠ABD + ∠ADB = 180°

⇒ 40° + 45° + ∠ADB = 180°

⇒ ∠ADB = 180° – 85° = 95°

and  ∠ADB + ∠ADC = 180°

⇒ ∠ADC = 180° – ∠ADB

= 180° – 95°

= 85°

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angle adb = 950

angle adc = 850

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ADB 95

ADC 85

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plz some one explain in steps plzzzzz

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