Question

15. In ∆ABC, B = 60°, C = 40°, AL parallel to BC and AD bisects A such that L and D lie on side BC. Find LAD. [Hint. BAL = 30°, CAL = 50° Then, BAD = DAC or 30° + LAD
= 50° - LAD ( AD bisects BAC)]

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Answer 1

Answer:

So angle AEC = 100

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Step-by-step explanation:

In triangle ABC

<A + <B+ <C = 180

60 + 40 + <C= 180

<C = 180 -100

<C = 80

since AE is the angle bisector of <BAC so <BAE + <EAC = 80 each 40 40

Now in triangle ABD <B+ <ADB + <BAD = 180 60 + 90 + X = 180 X = 180 150 = 30 so angle BAD = 30

Now BAD + DAE = 40 So DAE = 10

In triangle ADE 90 + 10 + X = 180

<D + <DAE + <AED = 180

X = 180 - 100 X = 80

so AED = 80

Now again in triangle AEC

<AEC + <EAC + <C = 180 X + 40 + 40 = 180

X = 180 - 80

X = 100

So angle AEC = 100

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Answer 2

Answer:

angle AEC =100°

hope it's correct and helpful for you