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3 years ago

∠2 and ∠3 form a linear pair. ∠3 and ∠1 are vertical angles. m∠1= 62 ∘ . What is m∠3?

2 answers:

8 0

For this problem, we do not need the first portion of the given information, so forget it for now. All we need to know is that 3 and 1 are vertical angles. Verticle angles are equal.
1=62
3 = 1
3 = 62
The measure of angle 3 is 62 degrees

Hope this helps =)

Rashid [163]3 years ago

3 0

M<3 is he sam as 123

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$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

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$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

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