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Mathematics 16 Online
OpenStudy (anonymous):

Find sin ð and cos β if 4, and y = 1

OpenStudy (anonymous):

here is triangle

jhonyy9 (jhonyy9):

so there is missed anything x wann being equale 4 ?

OpenStudy (anonymous):

explain

jhonyy9 (jhonyy9):

in the text of your exercise what is 4 ? there i think you have missed anything so we need calculing the sin $ and cos ß ,yes ?

OpenStudy (anonymous):

do you see an attatchment for the triangle

jhonyy9 (jhonyy9):

yes i see it but there are x and y like sides of this triangle so in the text of exercise you have wrote y=1 and whet is 4 ?

OpenStudy (anonymous):

x is 4

OpenStudy (anonymous):

i just need to know what sin and cos is?

jhonyy9 (jhonyy9):

ok suppose x=4 and you have wrote that y=1 i think that you have learned that sinus of an angle equale opposite side divide hypothenuse and cosinus of an angle is equal adjacent side divide hypothenuse do you know it ?

OpenStudy (anonymous):

no

OpenStudy (anonymous):

you kind of lost me at that sinus

jhonyy9 (jhonyy9):

so now i have wrote to you what you need using to calculi the needed cosinus and sinus of this angles ok ?

OpenStudy (anonymous):

ok

jhonyy9 (jhonyy9):

yes is sure that before all you need calculi the hypothenuse of this right triangle using pythagora right ?

OpenStudy (anonymous):

right

jhonyy9 (jhonyy9):

so can you solve it ?

OpenStudy (anonymous):

is it 17

jhonyy9 (jhonyy9):

what is 17 ?

OpenStudy (anonymous):

the hypotenuse

jhonyy9 (jhonyy9):

why is 17 ? can you write me here what say the teorem of pythagora ?

OpenStudy (anonymous):

nevermind

jhonyy9 (jhonyy9):

so let x and y two sides of an right triangle and the hypothenuse h pythagora said that h^2 = x^2 + y^2 ok ?

jhonyy9 (jhonyy9):

so than h= ?

jhonyy9 (jhonyy9):

are you here ?

jhonyy9 (jhonyy9):

so if hypothenuse squared equale 17 ,than this mean that h= sqrt17

jhonyy9 (jhonyy9):

than sin đ = opposite side / hypothenuse sin đ = y/sqrt17 = 1/ sqrt17 = sqrt17 / 17 after ratzionalize the denominator you get the sqrt17 /17 and for cos ß = adjacent side divide hypothenuse cos ß = y/sqrt17 = 1/sqrt17 = sqrt17 / 17

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