In ∆ ABC, √B=70• , √C= 60 • , BO AND CO ARE BISECTORS OF √ OF √ ABC AND √ACB RESPECTIVELY . THE MEASURE OF √ BOC IS OPTIONS 1) 105 °2) 90° 3) 115 ° 4) 125° prove it
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ABC is a triangle in which A=60°, bisectors of angles B and C meet at point O.
In triangle ABC
60°+B+C=180°. or. B+C=120°…………….(1)
B0 is bisector of. angle B and CO is bisector of angle C . Thus
angle OBC=B/2. , angle. OCB = C/2.
In triangle OBC
angle BOC+ angle OBC. + angle OCB. =180°
angle BOC. +B/2 +C/2=180°
angle BOC = 180°- 1/2(B+C). Putting B+C=120° from eqn.(1)
angle BOC. = 180°-1/2(120°) = 180°-60°. = 120°. Answer.
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In triangle ABC
60°+B+C=180°. or. B+C=120°…………….(1)
B0 is bisector of. angle B and CO is bisector of angle C . Thus
angle OBC=B/2. , angle. OCB = C/2.
In triangle OBC
angle BOC+ angle OBC. + angle OCB. =180°
angle BOC. +B/2 +C/2=180°
angle BOC = 180°- 1/2(B+C). Putting B+C=120° from eqn.(1)
angle BOC. = 180°-1/2(120°) = 180°-60°. = 120°. Answer.
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