The measure of angle x, in the given figure is:

35^o

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30^o

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45^o

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60^o

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Solution

The correct option is A
35^o

We know that the measure of an exterior angle of a triangle is equal to the sum of its two opposite interior angles.

So, x + 90^o = 125^o

Therefore, x = 35^o

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({sin}^{2} 30^{o} + {cos}^{2} 30^{o}) - ({sin}^{2} 60^{o} + {cos}^{2} 60^{o}) + ({sin}^{2} 45^{o} + {cos}^{2} 45^{o})

\frac{x {cosec}^{2} 30^{o} {sec}^{2} 45^{o}}{8 {cos}^{2} 45^{o} {sin}^{2} 90^{o}} = {tan}^{2} 60^{o} - {tan}^{2} 45^{o}

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\frac{1 + {tan}^{2} 30^{o}}{1 - {tan}^{2} 30^{o}} + {csc}^{2} 60^{0} - {cos}^{2} 45^{o} + {sin}^{2} 45^{o} + \frac{1 + {cot}^{2} 60^{o}}{1 - {cot}^{2} 60^{o}}

\frac{{tan}^{2} 60^{o} + 4 {sin}^{2} 45^{o} + {sin}^{2} 90^{o}}{3 {sec}^{2} 30^{o} + {csc}^{2} 60^{o} - {cot}^{2} 30^{o}}

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