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Darya [45]
3 years ago
14

Help please thank you

Mathematics
1 answer:
Leto [7]3 years ago
6 0
Ah, this my friend, is actually easier than it looks. I promise. Sort of. XD

Alright, so let's start with the basics. You have two shapes that look congruent, and obviously ARE congruent, but how they are congruent can be different.
HIJ (Shape 1) is congruently equal to (~=) shape LKJ (The order does indeed matter) by what?

Well, in terms of congruency, you have about 8 different ways, I only remember 4.
SSS (Side, Side, Side)
SAS (Side, Angle, Side)
ASA (Angle, Side, Side)
AAA (Angle, Angle, Angle)

This means that whichever it is, each must be identified as congruent to the other. If it's SAS, you must know, for certain (not you personally, you can guess, but that's not what they want, they want you to know based on the info they give you) that there are 2 sides that are congruent, and 1 angle that are congruent. Same for all the others, just plug and play. 

In the text, this question mentions that side HJ is congruently equal to JL. This means you have 1 set of sides identified as congruent. 
The text ALSO mentions that angle H is also congruently equal to angle L. This means you now have 1 set of angles that are congruently equal.

So far, you know you have 1 congruent set of sides (S) and One congruent set of angles (A)

Now, you also can see that based on what we already know, HIJ extends to LKJ, meaning the other angle would ALSO be congruent. 
This leaves you with "ASA" (Angle, Side, Angle), meaning 2 sets of angles are congruent, and 1 set of sides.

Your answer is A

~Hope this helps!
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Angle 0 is in the second quadrant, and sin 0=3/5
Aneli [31]
ANSWER


The value of theta in the second quadrant is,

143.1 \degree


EXPLANATION

The given trigonometric equation is
\sin( \theta)  =  \frac{3}{5}


This implies that,


\theta=  arcsin(\frac{3}{5} )


This implies that the principal solution is
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and it is an angle in the first quadrant.

But the solution we are looking for is an angle in the second quadrant.



This is given by
180 -  \theta = 180 - 36.9 = 143.1 \degree
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