I need help with geometry!

Mathematics

1 answer:

ss7ja [257]3 years ago

8 0

Step-by-step explanation:

If m is parallel to n and one of its angle is 25

Therefore, the vertically opposite angle to angle of 25 is also equal to 25

Now, we have to find x

Therefore, x=180-25 ( interer angles on the same side of the transversal so addition of both angles is 180)

Therefore, x =155

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If 1 is added to the numerator and to the denominator of a fraction,the value obtained is 4/5.If 5 is subtracted from it's numer

Step2247 [10]

Let n and d represent the numerator and denominator of the fraction.
.. (n +1)/(d +1) = 4/5
.. (n -5)/(d -5) = 1/2
Eliminating fractions from these equations gives
.. 5(n +1) = 4(d +1)
.. 2(n -5) = (d -5)
Subtracting the first equation from 4 times the second gives
.. 4(2(n -5)) -(5(n +1)) = 4(d -5) -(4(d +1))
.. 8n -40 -5n -5 = 4d -20 -4d -4
.. 3n = 21
.. n = 7
.. 2(7 -5) = d -5
.. 9 = d

The fraction is 7/9.

4 0

4 years ago

Given the general identity tan X =sin X/cos X , which equation relating the acute angles, A and C, of a right ∆ABC is true?

irakobra [83]

First, note that $m\angle A+m\angle C=90^{\circ}.$ Then

$m\angle A=90^{\circ}-m\angle C \text{ and } m\angle C=90^{\circ}-m\angle A.$

Consider all options:

$\tan A=\dfrac{\sin A}{\sin C}$

By the definition,

$\tan A=\dfrac{BC}{AB},\\ \\\sin A=\dfrac{BC}{AC},\\ \\\sin C=\dfrac{AB}{AC}.$

Now

$\dfrac{\sin A}{\sin C}=\dfrac{\dfrac{BC}{AC}}{\dfrac{AB}{AC}}=\dfrac{BC}{AB}=\tan A.$

Option A is true.

$\cos A=\dfrac{\tan (90^{\circ}-A)}{\sin (90^{\circ}-C)}.$

By the definition,

$\cos A=\dfrac{AB}{AC},\\ \\\tan (90^{\circ}-A)=\dfrac{\sin(90^{\circ}-A)}{\cos(90^{\circ}-A)}=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{AB}{AC}}{\dfrac{BC}{AC}}=\dfrac{AB}{BC},\\ \\\sin (90^{\circ}-C)=\sin A=\dfrac{BC}{AC}.$