Question

EXERCISE 11 (A) 1. In the figure, drawn at the right AB BC, AC = AE. CE = AE, AF = EF. <ABC-90, <CDE -90​

Answer 1

Answer:

m∠EBC=65.25°

AC=20 in.

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.

similarly in right angled triangle ΔBEC we have:

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

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

Hence side AC=20 in.

Step-by-step explanation:

Hope this helps:)  ジョセフ