FCDEF is congruent to MNPQ, then angle P is congruent to

Mathematics

1 answer:

SpyIntel [72]2 years ago

7 0

Answer:

The angles of both figures are listed in the same order. The first angle of the first one is congruent to the first angle of the second, and the second angle of the first one is congruent to the second angle of the second figure, and so on.

Step-by-step explanation:

You have a repeated 'F' in naming the first figure, so I think you made an error. But the rule in my answer is true and you can apply it.

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Find the exact value.

levacccp [35]

$\text{Given that,}\\\\\cot A = -\dfrac{9}{\sqrt{40}}\\\\\\\implies \tan A = -\dfrac{\sqrt{40}}9\\\\\\\implies \tan^2 A = \dfrac{40}{81}\\\\\\\implies \sec^2 A -1 = \dfrac{40}{81}\\\\\implies \sec^2 A = 1 + \dfrac{40}{81}\\\\\implies \sec^2 A = \dfrac{121}{81}\\\\\implies \sec A = \pm \sqrt{\dfrac{121}{81}}= \pm\dfrac{11}{9}\\\\\\\text{Since the angle A is in quadrant II, the value of sec A will be negative.}\\\\\\\text{Hence,}~ \sec A=-\dfrac{11}{9}$