Question

Given the triangle below, what is sec ∡B?

Answer 1

i'llk type it in the calculater

Answer 2

so you need to solve for adjacent side first
\[AB = \sqrt{7^2 - 4^2}\]

Answer 3

ugh i typed it wrong again =_=

sec = hypotenuse/adjacent

Answer 4

wait so what do i put in the calculater?

Answer 5

\[\sqrt{7^2 - 4^2}\]

Answer 6

soleve that first then tell me the answer

Answer 7

note: retain the sqrt symbol

Answer 8

i got 5.74456264653803

Answer 9

i said retain the sqrt symbol so just solve 7^2 - 4^2

Answer 10

ok

Answer 11

33

Answer 12

so the adjacent side is \(\sqrt{33}\) now do hypotenuse/adjacent

Answer 13

ok

Answer 14

isn't it the same 7^-4^ ?

Answer 15

no

Answer 16

hypotenuse over adjacent is \[\frac{7}{\sqrt{33}}\]

Answer 17

7 is hypotenuse
\(\sqrt{33}\) is adjacent

Answer 18

do i put everything over 33?

Answer 19

no you rationalize

\[\frac{7}{\sqrt{33}} \times \frac{\sqrt{33}}{\sqrt{33}}\]

Answer 20

ok so its going to be