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Given:

AB

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

Given: AB ≅ BC and AE = 10 in m∠FEC = 90° m∠ABC = 130°30’ Find: m∠EBC, AC-example-1
User Ivonet
by
7.3k points

2 Answers

1 vote

Answer:

65.25

20in

Explanation:

User Eriknoyes
by
7.4k points
2 votes

Answer:

  • m∠EBC=65.25°
  • AC=20 in.

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.


\tan 24.75=(BE)/(AE)=(BE)/(10)\\\\BE=10\tan 24.75-------(2)

similarly in right angled triangle ΔBEC we have:


\tan 24.75=(BE)/(EC)\\\\\tan 24.75=(BE)/(EC)\\\\EC=(BE)/(\tan 24.75)------(3)

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


EC=10 \text{in.}

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

Hence side AC=20 in.


User Dbrrt
by
7.4k points