Use the information given in the following figure to find:
(1) x
(2) $∠ B$ and $∠ C$

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Solution

Here, given that,
$∠ A = 90^{\circ} ∠ B = (2 x + 4)^{\circ} ∠ C = (3 x - 5)^{\circ} ∠ D = (8 x - 15)^{\circ}$
We know that,
Sum of all the interior angles in a quadrilateral is $360^{\circ}$
So,
$∠ A + ∠ B + ∠ C + ∠ D = 360^{\circ}$
$90^{\circ} + (2 x + 4^{\circ}) + (3 x - 5^{\circ}) + (8 x - 15^{\circ}) = 360^{\circ}$
On further calculation, we get
$90^{\circ} + 2 x + 4^{\circ} + 3 x - 5^{\circ} + 8 x - 15^{\circ} = 360^{\circ} 74^{\circ} + 13 x = 360^{\circ} 13 x = 360^{\circ} - 74^{\circ} 13 x = 286^{\circ} x = {(\frac{286}{13})}^{\circ} We get, x = 22^{\circ} The value of x i s 22^{\circ} Now, ∠ B = 2 x + 4^{\circ} = 2 \times 22^{\circ} + 4^{\circ} = 48^{\circ} ∠ C = 3 x - 5^{\circ} = 3 \times 22^{\circ} - 5^{\circ} = 61^{\circ}$
Therefore, $∠ B = 48^{\circ}$ and $∠ C = 61^{\circ}$

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