In a quadrilateral ABCD, $∠ A = 90^{\circ}$ , $∠ B = 3 x$ , $∠ C = 3 x + 5$ , $∠ D = 8 x - 15$ .
Then, the value of $∠ D$ is

$65^{\circ}$

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$60^{\circ}$

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$20^{\circ}$

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$145^{\circ}$

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Solution

The correct option is D
$145^{\circ}$

Since the sum of interior angles of a quadrilateral is $360^{\circ}$
$90 + 3 x + 3 x + 5 + 8 x - 15 = 360 90 + 14 x - 10 = 360 14 x = 360 - 80 x = \frac{280}{14} = 20 T h e n, ∠ D = 8 x - 15$
= $8 (20) - 15 = 145^{\circ}$

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In a quadrilateral ABCD, ∠A=90∘, ∠B=3x, ∠C=3x+5, ∠D=8x−15. Then, the value of ∠D is

The correct option is D 145∘ Since the sum of interior angles of a quadrilateral is 360∘ 90+3x+3x+5+8x−15=36090+14x−10=36014x=360−80x=28014=20Then,∠D=8x−15 = 8(20)−15=145∘

In a quadrilateral ABCD, $∠ A = 90^{\circ}$ , $∠ B = 3 x$ , $∠ C = 3 x + 5$ , $∠ D = 8 x - 15$ .
Then, the value of $∠ D$ is

The correct option is D
$145^{\circ}$

Since the sum of interior angles of a quadrilateral is $360^{\circ}$
$90 + 3 x + 3 x + 5 + 8 x - 15 = 360 90 + 14 x - 10 = 360 14 x = 360 - 80 x = \frac{280}{14} = 20 T h e n, ∠ D = 8 x - 15$
= $8 (20) - 15 = 145^{\circ}$