in figure sides AB and BC of ∆ABC are produced to point E and D respectively if ∆ABD=120° and ∆EAC=110° then find ∆ACB
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Step-by-step explanation:
QIn ∆ABC angle C is obtuse angle , therefore A+B+C=180°…………….(1)
The bisectors of exterior angles at A and B meet BC and AC produced at
D and E respectively.
In ∆ ABD , AB=AD ( given) therefore angle ADB= angle ABD=angle B.
Angle BAD= A+(90-A/2)=90+A/2.
Angle BAD+angle ABD+angle ADB =180°
(90°+A/2)+B+B = 180
or. A/2+2B=90.
or. A+4B=180°……………(2)
In ∆ ABE , AB=BE ( given) therefore angle BEA=angle BAE= angle A.
Angle ABE=B+(90°-B/2)= 90°+B/2.
Angle ABE+angleBAE+angle BEA=180°
(90°+B/2)+A+A=180°
or. 2A+B/2=90°
or. 4A+B=180°……………..(3)
by solving eqn. (2) and (3) A=B=36°. , putting A=B=36° in eqn. (1)
36°+36°+C=180°
or. C=180°-72° = 108°
Thus , angle ACB= 108°. Answer.
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