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In the given figure, O is the centre of the circle and BA = AC. If `angleABC=50^@`, find `angleBOC` and `angleBDC.` |
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Answer» BA=AC `/_ABC=/_ACB` `50^@=/_ACB` `In/_ABC` `/_A+/_B+/_C=180^@` `/_A+50+50=180` `/_A=80^@` `/_BOC=2/_A` `/_BOC=2*80=160^@` `/_BAC+/_BDC=180` `80+/_BDC=180` `/_BDC=100`. |
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