Question

Can someone help me with this question?

17. what is the measure of angle B in the triangle below?
a. 21.0 degrees
b. 22.6 degrees
c. 42.7 degrees
d. 67.4 degrees

Answer 1

triangle image?

Answer 2

Could someone help explain it also, because I honestly don't know how to do it.

Answer 3

well you need use a trig ratio.... it doesn't matter which ratio you select as they will all give the same answer...

any choices..?

Answer 4

I don't know what you mean.

Answer 5

have you studied trigonometry...?

Answer 6

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Answer 7

do you know what the "opposite side" for angle B is in that triangle?

Answer 8

anyhow.... all you'd need is to recall your SOH CAH TOA \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\)

Answer 9

I know that, but it isn't the side that we are trying to find, it's the angle.

Answer 10

right so.. ahemm
say... do you know what the "opposite side" of angle B is?

Answer 11

keep in mind, an opposite side, is the one "facing off" the angle
so if you put your "eye" on that angle, the side you'd be facing will be the "opposite side"

so.... which is it from angle B then?

Answer 12

the left side

Answer 13

hello?

Answer 14

in this case, yes, or side 5
thus so that means
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin^{-1}[sin(\theta)]=sin^{-1}\left( \cfrac{opposite}{hypotenuse} \right)
\\ \quad \\
\theta=sin^{-1}\left( \cfrac{opposite}{hypotenuse} \right)

\\ \quad \\

% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}\implies \implies cos^{-1}[cos(\theta)]=cos^{-1}\left( \cfrac{adjacent}{hypotenuse} \right)
\\ \quad \\
\theta=cos^{-1}\left( \cfrac{adjacent}{hypotenuse} \right)

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\implies \implies tan^{-1}[tan(\theta)]=tan^{-1}\left( \cfrac{opposite}{adjacent} \right)
\\ \quad \\
\theta=tan^{-1}\left( \cfrac{opposite}{adjacent} \right)\)

Answer 15

to find the degree of the angle?

Answer 16

yes, you could use any of them really
because you're given all sides

so use either, sine or cosine or tangent

Answer 17

that still doesn't make sense to me. there should be a certain formula to find the degree of the angle. you confused me by putting all of that mumbo jumbo up there.

Answer 18

well... have you covered \(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad \qquad
% cosine
cos(\theta)=\cfrac{adjacent}{hypotenuse}

\\ \quad \\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\) yet?

Answer 19

yes, I already said that. could you explain to me how to find the degree of any triangle using those??

Answer 20

welll sure, let us use say the sine then
\(\bf sin(\theta)=\cfrac{opposite}{hypotenuse}\implies sin^{-1}[sin(\theta)]=sin^{-1}\left( \cfrac{opposite}{hypotenuse} \right)
\\ \quad \\
\theta=sin^{-1}\left( \cfrac{opposite}{hypotenuse} \right)\)

the \(\Large sin^{-1}\) should be in your calculator, so just plug in the values for opposite and hypotenuse

make sure your calculator is in Degree mode

Answer 21

opposite is 5
hypotenuse 15
thus
\(\bf sin(B)=\cfrac{5}{13}\implies sin^{-1}[sin(B)]=sin^{-1}\left( \cfrac{5}{13} \right)
\\ \quad \\
\measuredangle B=sin^{-1}\left( \cfrac{5}{13} \right)\)

Answer 22

hmm
opposite is 5
hypotenuse 13 anyway

Answer 23

okay I got it now. you weren't explaining it very well at first.

Answer 24

heheh, ok