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Mathematics 12 Online
OpenStudy (anonymous):

Given the triangle below, what is sec ∡B?

OpenStudy (anonymous):

OpenStudy (anonymous):

i'llk type it in the calculater

OpenStudy (lgbasallote):

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

OpenStudy (lgbasallote):

ugh i typed it wrong again =_= sec = hypotenuse/adjacent

OpenStudy (anonymous):

wait so what do i put in the calculater?

OpenStudy (lgbasallote):

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

OpenStudy (lgbasallote):

soleve that first then tell me the answer

OpenStudy (lgbasallote):

note: retain the sqrt symbol

OpenStudy (anonymous):

i got 5.74456264653803

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

33

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

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

OpenStudy (lgbasallote):

no

OpenStudy (lgbasallote):

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

OpenStudy (lgbasallote):

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

OpenStudy (anonymous):

do i put everything over 33?

OpenStudy (lgbasallote):

no you rationalize \[\frac{7}{\sqrt{33}} \times \frac{\sqrt{33}}{\sqrt{33}}\]

OpenStudy (anonymous):

ok so its going to be|dw:1345362932506:dw|

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