Question

sec∠C - cot∠A = 3x

fraction 7 minus square root of 13 over 18

fraction 6 minus square root of 13 over 7

fraction square root of 13 over 3

square root of 13 over 3

Answer 1

Please help me figure this out!

Answer 2

Okay first lets find the missing side of the triangle with the Pythagorean theorem

Answer 3

a^2+b^2=c^2 ??

Answer 4

yep and were looking for b. a=6 and c=7

Answer 5

36+b^2=49
b^2=13
and b=sqrt(13)
with me so far?

Answer 6

B=13 :)

Answer 7

don forget the square root cuz b^2=13 not b

Answer 8

Oh yeah oops.

Answer 9

No that we know all the sides of the triangle we can get into the trigonometry. Lets just change it all to sines and cosines cuz those are the easy ones

Answer 10

Alright :)

Answer 11

so what is sec in terms of sin and cos?

Answer 12

Sec < = Hypotenuse/Adjacent

Answer 13

Okay that works. its the secant of C so whats adjacent to C and whats the hypotenuse

Answer 14

Adjacent is square root of 13 and Hypotenuse is 7 right?

Answer 15

Adjacent to angle C is 6 cuz sqrt(13) is segment AB which is opposite to C but the hypotenuse is 7 so sec (C)=hyp/adj=7/6 got it?

Answer 16

Ohh yeah I get those mixed up your right haha okay I got it. :)

Answer 17

Now move on to cotangent of A. First, what's cotangent?

Answer 18

Cotangent Adjacent/Opposite so therefore Adjacent 13 and Opposite 6 right?

Answer 19

??

Answer 20

Sorry. I had to eat dinner. sqrt(13) not 13 but yes thats right though.

Answer 21

so you end up with 7/6-sqrt(13)/6=3x got it?

Answer 22

Ohh okay! 6X3=18 so 7- sqr. of 13 over 18??

Answer 23

Yah exactly!

Answer 24

Wow thanks ! haha I get it now.

Answer 25

No problem!