Question

in the given figureOA=3m,OB=4 cm ,angleAOV= 90 degree and AC=12 and BC= 13 cm. then prove that angleCAB= 90 degree.

Answer 1

AOB = 90°
${ab}^{2} = {oa}^{2} + {ob}^{2}$
ab^2=9+16=25cm
ab=5
${bc}^{2} = {ab}^{2} + {ac}^{2}$
${ab}^{2} = {bc}^{2} - {ac}^{2}$
=169-144
=25
ab=5
CAB=90°