Given: AB ≅ BC , AE ≅ FC Prove: m∠AEC=m∠AFC

Answer:

EC/AB = FC/MA

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 equalt

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=49.5

∠BAE=24.75°  (DIVIDE BOTH SIDE BY 2)

Now in triangle ΔBEC  we have:

∠BEC=90° , ∠BCE=24.75°

SO,

∠BEC+∠BCE+∠EBC=180°.

Hence, 90+24.75+∠EBC=180

∠EBC=180-(90+24.75)

∠EBC=65.25°

Now we are given AE = 10 in

Also ∠BEA= 90°.

And ∠BAE=24.75°; hence using trignometric identity to find the measure of side BE.

\begin{gathered}\tan 24.75=\dfrac{BE}{AE}=\dfrac{BE}{10}\\\\BE=10\tan 24.75-------(2)\end{gathered}tan24.75=AEBE=10BEBE=10tan24.75−−−−−−−(2)

similarly in right angled triangle ΔBEC we have:

\begin{gathered}\tan 24.75=\dfrac{BE}{EC}\\\\\tan 24.75=\dfrac{BE}{EC}\\\\EC=\dfrac{BE}{\tan 24.75}------(3)\end{gathered}tan24.75=ECBEtan24.75=ECBEEC=tan24.75BE−−−−−−(3)

Hence, using equation (2) in equation (3) we get:

EC=10 \text{in.}EC=10in.

Hence AC=AE+EC=10+10=20 in.

Hence side AC=20 in.

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