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Question

In quadrilateral $$ABCD$$, if $$\angle A = 60^{\circ}$$ and $$\angle B : \angle C : \angle D = 2 : 3 : 7$$, then find $$\angle D$$.


A
175
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B
135
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C
150
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D
120
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Solution

The correct option is D $$175^{\circ}$$
In quadrilateral $$ABCD$$, $$\angle A$$ = $$60^\circ$$ and $$\angle B: \angle C: \angle D = 2 : 3 : 7  $$        [Given].

Let $$\angle B, \angle C, \angle D$$ are $$2x^\circ$$ , $$3x^\circ$$ and $$7x^\circ$$.

$$\Rightarrow$$  $$\angle A+ \angle B+ \angle C + \angle D $$=$$ 360^\circ$$    [Sum of all angles of quadrilateral is $$360$$$$^\circ$$]

$$\Rightarrow$$  $$60^\circ + 2x^\circ+ 3x^\circ +7x^\circ$$ = $$360^\circ$$

$$\Rightarrow$$  $$12x^\circ$$ = $$360^\circ - 60^\circ$$

$$\Rightarrow$$  $$x^\circ$$ = $$\dfrac {300^\circ}{12}$$

$$\Rightarrow$$  $$x^\circ$$ = $$25^\circ$$.

Now, $$\angle D$$=$$7x^\circ$$.

Substitute value of $$x$$ we get,
$$\Rightarrow$$  $$\angle D$$=$$ 7\times 25^o$$

$$\therefore$$  $$\angle D = 175^\circ$$.

Hence, option $$A$$ is correct.

Mathematics

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