Find the sum of $∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 + ∠ 9$

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Solution

Since, sum of all angles of triangle is $180 °$
We have,
In $△ A B C$
$\Rightarrow ∠ 1 + ∠ 2 + ∠ 3 = 180 °$
And in $△ A C D$
$\Rightarrow ∠ 4 + ∠ 5 + ∠ 6 = 180 °$
And in $△ A E D$
$\Rightarrow ∠ 7 + ∠ 8 + ∠ 9 = 180 °$
$∴ ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 + ∠ 9 = 3 ✖ 180 ° = 540 °$
Hence, the sum of $∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 + ∠ 9 = 540 °$ .

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Find the sum of ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8+∠9

Since, sum of all angles of triangle is 180°We have,In △ABC⇒∠1+∠2+∠3=180°And in △ACD⇒∠4+∠5+∠6=180°And in △AED⇒∠7+∠8+∠9=180°∴∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8+∠9=3✖180°=540°Hence, the sum of ∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8+∠9=540°.

Find the sum of $∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 + ∠ 9$