Please solve q 74 ?

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Please solve q 74 ?

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Answer 1 · 2018-05-13T19:34:40

$50^{\circ}$ $50^@$

Explanation:

$B ˆ D C = 150^{\circ}$ $BhatDC=150^@$ so the reflex angle is $210^{\circ}$ $210^@$
Looking at the quad BGCD you have $100^{\circ} + 210^{\circ} + \frac{B}{2} + \frac{C}{2} = 360^{\circ}$ $100^@+210^@+B/2+C/2=360^@$
So $\frac{B}{2} + \frac{C}{2} = 50^{\circ}$ $B/2+C/2=50^@$

Triangle BGF has $80^{\circ} + \frac{B}{2} + B ˆ F G = 180^{\circ}$ $80^@+B/2+BhatFG=180^@$
$A ˆ F G = 80^{\circ} + \frac{B}{2}$ $AhatFG = 80^@+B/2$ (exterior angle on a triangle=sum of opposite two interior angles)

Triangle CEG has $80^{\circ} + \frac{C}{2} + C ˆ G E = 180^{\circ}$ $80^@+C/2+ChatGE=180^@$
$A ˆ E G = 80^{\circ} + \frac{C}{2}$ $AhatEG= 80^@+C/2$ (exterior angle on a triangle=sum of opposite two interior angles)

So the quad AEGF has
$100^{\circ} + (80 + \frac{B}{2}) + (80 + \frac{C}{2}) + F ˆ A E = 360^{\circ}$ $100^@+(80+B/2)+(80+C/2)+FhatAE=360^@$

$260^{\circ} + \frac{B}{2} + \frac{C}{2} + F ˆ A E = 360^{\circ}$ $260^@+B/2+C/2+FhatAE=360^@$

$\frac{B}{2} + \frac{C}{2} + F ˆ A E = 100$ $B/2+C/2+FhatAE=100$

But we know $\frac{B}{2} + \frac{C}{2} = 50^{\circ}$ $B/2+C/2=50^@$ so $F ˆ A E = 50^{\circ}$ $FhatAE =50^@$

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Please solve q 74 ?

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