Given that PQ and RS are drawn with KL as tranversal intersecting PQ at M and RS at point N. Angle QMN is congruent to angle LNS because they are alternate to each other. The theorem that Kari can use to show that the meansure of QML is supplementary to the measure of angle SNK is Alternate Exterior Angles Theorem.
This is because angle KNR is equal to QML by alternate exterior angles theorem so is angle MLP and SNK
-10a+15b + 15b-10a = -20a+30b
Answer:
It's the second and third one!!
Step-by-step explanation:
Answer:
If the sine of an angle equals to {2}/{5}, then the tangent of the angle could be:
<em>5.
</em>
Step-by-step explanation:
- The sine of an angle is equal to the length of the opposite leg over the hypotenuse.
- <em>HINT: SOHCAHTOA</em>
<u>To find the adjacent leg (a): </u>
Using the pythagorean theorem: 
- <em>a </em>and <em>b</em> are the lenght of the sides and <em>c </em>is the hypotenuse
we know one side and the hypotenuse. Therefore plugging into the formula we get:

where <em>a</em> is the adjacent leg of the triangle, solving for a we get:



<u>To find the tangent of the angle: </u>
Since tangent is opposite leg over the adjacent leg:
tan θ = 