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3 years ago

Given m∠D = 70°, m∠E = 42, and m∠EHG = 68°, prove DF is parallel to GH.

Mathematics

1 answer:

Pie3 years ago

4 0

we know that

If DF is parallel to GH

then

the measure of the angle EGH must be equal to 70° by corresponding angles

<u>Find the measure of the angle EGH </u>

In the triangle EGH

the sum of the internal angles must be equal to 180 degrees

42°+68°+∠EGH=180°

solve for ∠EGH

∠EGH=180-(42+68)

∠EGH=180-110

∠EGH=70°

therefore

The lines DF and GH are parallel

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3 years ago

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Vera_Pavlovna [14]

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$tan\:\left(\frac{5\pi }{6}\right)=tan\left(-\frac{\pi }{6}\right)$

Option A is correct.

Step-by-step explanation:

We need to find equivalent of $tan\:\left(\frac{5\pi }{6}\right)$

First we solve $tan\:\left(\frac{5\pi }{6}\right)$

We get $-\frac{1}{\sqrt{3} }$

Now checking all the options.

Option A: $tan\:(-\frac{\pi}{6} )$

Solving $tan\left(-\frac{\pi }{6}\right)\: we\: get\: \mathbf{ -\frac{1}{\sqrt{3} }}$

Option B: $tan\left(\frac{7\pi }{6}\right)$

Solving $tan\left(\frac{7\pi }{6}\right)\: we\: get\: \mathbf{\frac{1}{\sqrt{3} }}$

Option C: $cot\left(\frac{5\pi }{6}\right)$

Solving $cot\left(\frac{5\pi }{6}\right)\:we\:get\:\mathbf{-\sqrt{3} }$

Option D : $tan\left(-\frac{5\pi }{6}\right)$

Solving $tan\left(-\frac{5\pi }{6}\right) \:we\:get:\mathbf{\frac{1}{\sqrt{3} } }$

So, looking at the options, only Option A has the same result as given question

So, $tan\:\left(\frac{5\pi }{6}\right)=tan\left(-\frac{\pi }{6}\right)$

Option A is correct.

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2 years ago