Given: Circles k1(O, r=OK) and k2(O, r=OS) -concentric KE tangent to k2(O, r=OS) at S Prove: KS ≅ SE

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.

step-by-step explanation: