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4. In the given figure, AOB is a diameter of the cirdle with centre O and AC is a tangent to the circle at A.If ZBOC130°, then find ZACO130° |
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Answer» AOC + COB = 180 ( ANGLES IN A STRAIGHT LINE )AOC = 180 - COBAOC = 180 - 130AOC = 50AO = CO ( RADIUS OF A CIRCLE)OAC = ACO = X ( BY LAW OF ISOSCELES TRIANGLE)OAC + ACO + COA = 180 ( SUM OF ANGLE OF TRIANGLE)X + X + 50 = 1802X + 50 = 1802X = 180 - 502X = 130X = 65THEREFORE , ACO = 65 Like my answer if you find it useful! given: angle boc=90°proof: angle BOC+ angle COA=180°130°+ ANGLE COA=180°ANGLE COA=180°-130°THEREFORE, ANGLE COA =50°ANGLE OAC=90° by thereorm 10.1Angle AOC+COA+OAC=180°( Triangle sum property)AOC+50°+90°=180°AOC+140°=180°AOC=40° sorry one mistakeAOC+ACO+OAC=180°50°+ACO+90°=180°ACO=180°-140°angle ACO=40° |
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