In $△ A B C$ , as shown in the following figure, side BC is extended through C to D. $m ∠ A C D = 6 x + 40, m ∠ B A C = 3 x + 30 a n d m ∠ A B C = 4 x + 5$ find $m ∠ A C B$ .

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Solution

By exterior $∠$ property of $Δ$
$∠ A B C + ∠ B A C = ∠ A C D$
$\Rightarrow 4 x + 5 + 3 x + 30 = 6 x + 40$
$\Rightarrow x = 5$

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