pls help me?

Question

my teacher is soooo bad he gave us a problem ... and he didn't even explain it ..... pls help me ..pls

here is the problem

1. sec(-2A)= csc 20 A

solve for A

pls help me

Answer 1

do your own math... lazy

Answer 2

Ans.: A=5 degrees

For a given right triangle, since sec and csc are co functions, their functions can only be equal if the sum of their respective angles representing the two acute angles is 90 degrees(meaning their angles are complementary angles).

solution:

-2A + 20A = 90
18A =90
A= 90/18
A=5 degrees

Answer 3

A equals 1

Answer 4

First, start off with the definition of sec and csc. sec (theta) = csc (pi/2 - theta). Therefore, you know that pi/2 + 2A = 20A. Use algebra to solve the rest.

Answer 5

This is a piece of cake:
sec(-2A)= 1/ (cos-2A) = 1/ cos2A
cosec 20A= 1/ sin20A

so cos2A= sin20A

and since cos (90-X) = sin X, so sin20A = cos (90-20A)
and then cos 2A= cos (90-20A)

2A = 90-20A
22A=90, so A will be 90/22= 4.09 approximately

Answer 6

Ah.....I remember problems like these. Hmmm.....you must change cosecant into a secant using the rules of tirg, then you can can cancel one of the two A's and use inverse trig to find the angle A.


O yeah..... YAHOO IS NOT MEANT FOR HOMEWROK

Answer 7

Hey its the proper solution

sec(-2A)= sec(2A)

sec(2A)=csc20A
cos2A=sin20A
cos2a=cos(Pie/2-20A)
pie/2-20A= 2n(pie) +-(both + and -) 2A
Now u can easily manupulate taking + sign and then - sign.

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Answer 8

If i knew how to solve it i would have surely helped you!! But sorry, i cant!!!! Hate to say No to a girl like you!!!

Answer 9

I don't think YA is meant to help you with mundane maths homework.

Answer 10

Sorry