Given:AB ≅BC and AE = 10 in m∠FEC = 90° m∠ABC = 130°30’ Find: m∠EBC, AC

Answer:m∠EBC = 65.25°AC=20 in.Step-by-step explanation:We are given AB ≅ BC that means that side AB and side BC are equal also we know that angle opposite to equal sides are equal.Hence, ∠BAE=∠BCE-------(1)Also ∠AEB=∠CEB.Now we are given that: ∠ABC = 130°30’ i.e. in degrees it could be given as:60'=1°30'=(1/2)°=0.5°Hence ∠ABC = 130°30’=130+0.5=130.5°Also we know that sum of all the angles in a triangle is equal to 180°.Hence,∠BAE+∠BCE+∠ABC=180°.2∠BAE+130.5=180 (using equation (1))2∠BAE=49.5Dividing both sides by 2 we get;∠BAE=24.75°Now in triangle ΔBEC  we have:∠BEC=90° , ∠BCE=24.75°SO,∠BEC+∠BCE+∠EBC=180°.Hence, [tex]90+24.75+ \angle EBC=180[/tex]∠EBC=[tex]180-(90+24.75)[/tex]∠EBC=65.25°Now we are given AE = 10 inAlso ∠BEA= 90°.And ∠BAE=24.75°; hence using trigonometric identity to find the measure of side BE.[tex]tan24.75=\frac{BE}{AE} = \frac{BE}{10}\\\\BE= 10 \ tan24.75 \ \ \ \ \ equation \ 2[/tex]similarly in right angled triangle ΔBEC we have:[tex]tan24.75=\frac{BE}{EC}\\\\EC=\frac{BE}{tan24.75} \ \ \ \ \ \ \ \ \ equation \ 3[/tex]Hence, using equation (2) in equation (3) we get:[tex]EC = \frac{10 \ tan24.75}{tan24.75} =10in[/tex]Hence AC=AE+EC=10+10=20 in.Hence side AC=20 in.