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let us angle ABO=1,angle OBC=2,angle OCA=3 and angle BCO=4
GIVEN. 1=2 ,3=4
RTP. BOC=90+1/2 A
PROOF. In ∆BOC
2+BOC+4=180°
1/2 B+BOC+1/2 C=180°
1/2(B+C)+BOC=180°
(since, A+B+C=180°
therefore,B+C=180°-A)
1/2(180-A)+BOC=180°
90°-1/2 A+BOC=180°
BOC=180°-90°+1/2A
BOC=90°+1/2 A
GIVEN. 1=2 ,3=4
RTP. BOC=90+1/2 A
PROOF. In ∆BOC
2+BOC+4=180°
1/2 B+BOC+1/2 C=180°
1/2(B+C)+BOC=180°
(since, A+B+C=180°
therefore,B+C=180°-A)
1/2(180-A)+BOC=180°
90°-1/2 A+BOC=180°
BOC=180°-90°+1/2A
BOC=90°+1/2 A
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