Question

in triangle abc angle B =45 degree angle c = 55 degree and bisector of angle A meets BC at a point O find angle ADB and angle ADC

Answer 1

In the triangle ADB sum of other angles is=45+40=85 so third angle will beADB=95 and ADC=95

Answer 2

** CORRECTION IN QUESTION :- bisector of angle A meets BC at a point D find angle ADB and angle ADC, ( it will be point D , not O)

According the question :-

∠B=45,

∠C=55

First we have to find out ∠A

In Δ ABC sum of three  angles is 180°

So,  ∠A + ∠B + ∠C = 180°

∠A + 45° + 55° = 180°

∠A=180°−(45°+55°)

=80°

∠A =∠ BAD +∠ CAD= 80°

AD is the bisector of ∠A.

∴ ∠ BAD =∠ CAD

Hence ∠ BAD = $\frac{1}{2}$ ∠A= $\frac{80}{2}$= 40° = ∠ CAD

Now we have to find out ∠ADB.

In △ABD:-

∠ BAD + ∠ ABD + ∠ ADB = 180°

⇒40°+45°+∠ADB=180°

⇒∠ADB= 180° - ( 40° + 45°)

            =95°

Now we have to find out ∠ADC.

In△ADC:-

⇒∠CAD + ∠ACD + ∠ADC= 180°

⇒40°+55°+∠ADC=180°

⇒∠ADC= 180° - ( 40° + 55°)

             =85°.

Ans :-  ∠ADB=95° and ∠ADC=85°.

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