Question

Triangle LMN has vertexes at L(-1, -6), M(1, -6), and N(1, 1). Find the measure of angle L to the nearest degree.
(Hint: Sketch the triangle on the coordinate plane and find the side lengths using the Distance Formula first.)

Answer 1

k

Answer 2

I found the other length for LN and it would be 7.28

Answer 3

MN is 7 and LM is 2.

Answer 4

ok
yes, I'd say, get the segment's length first, then use the law of cosines

\(\bf \cfrac{{\color{blue}{ a}}^2+{\color{red}{ b}}^2-c^2}{2{\color{blue}{ a}}{\color{red}{ b}}}=cos(C)\implies cos^{-1}\left(\cfrac{{\color{blue}{ a}}^2+{\color{red}{ b}}^2-c^2}{2{\color{blue}{ a}}{\color{red}{ b}}}\right)=\measuredangle C\)

Answer 5

I think you miscalculated the LM side length, it is for sure larger than 2

Answer 6

How would it not be 2. Like i did the distance formula and used the graph and its 2 lol

Answer 7

sorry, i see

Answer 8

my mistake

Answer 9

so you have...

Answer 10

and 7.28

Answer 11

it is actually a right triangle,

Answer 12

yes

Answer 13

MN is vertical, have same x coordinate and LM is horizontal with the same y coordinate

Answer 14

you can use whichever trig function now since it is a right triangle

Answer 15

tangent of L is the opposite side over the adjacent side

Answer 16

Tan(L) = 7/2

Answer 17

it was actually pretty easy, no law of cosines to calculate...just my plotting points wrong to start

Answer 18

But how do you get L?

Answer 19

have you used the arctangent function? tan^(-1) ...usually 2nd or shift and tangent on the calculator

Answer 20

so it'd be 74.1?

Answer 21

looks about right

Answer 22

And N would be 16?

Answer 23

tan^(-1)(7/2) =

Answer 24

yeah the three angles total 180

Answer 25

Thanks !

Answer 26

no prob, welcome