Question

solve (x-h)^2+B^2=r^2......solve for B

Answer 1

\[(x-h)^2+B^2=r^2\]

Answer 2

start by subtracting \(\Large (x-h)^2\) from both sides

Answer 3

after you isolate \(\Large B^2\) you can take the square root of both sides to isolate B itself.

Answer 4

so i got \[B=\sqrt{r^2-x-h^2}\]

Answer 5

it should be \[\Large B = \sqrt{r^2 - (x-h)^2}\] assuming B is nonnegative

Answer 6

oh ok. can you help me out by testing on some inverse trig functions?

Answer 7

post your question

Answer 8

like give me questons similar to arcsin(tang1/2).

Answer 9

ok how about `cos(arcsin(1/2))`

Answer 10

ok. umm.. sqrt3/2?

Answer 11

yes

Answer 12

give me a few more. particularly where the range changes cause those confuse me.

Answer 13

arcsin(cos(5pi/4))

Answer 14

hmmm i have no clue.

Answer 15

find the value of cos(5pi/4) first

Answer 16

thats -sqrt2/2

Answer 17

now find out what value of theta makes sin(theta) = -sqrt(2)/2 true

keep in mind that the range of arcsine is -pi/2 <= y <= pi/2

Answer 18

so would it be pi/4?

Answer 19

no

Answer 20

-pi/4??

Answer 21

correct

Answer 22

2 more please.

Answer 23

im trying to figure out sec(arcsin(x-1))

Answer 24

what is b equal to in terms of x

Answer 25

im writing an algebraic expression

Answer 26

im solving for b.

Answer 27

so (x-1)^2+B2=1....subtract (x-1)^2 on both sides.....then im left with B^2=1-(x-1)^2..... i square both sides..... now i have B= sqrt1-(x-1)^2

Answer 28

yes \[\Large b = \sqrt{1-(x-1)^2}\]

Answer 29

and now im suppose to relate that as a algebraic expression. so back to sec(arcsin(x-1))

Answer 30

since im working inside out, sin is opp/hyp. so x-1/sqrt1-(x-1). but i dont know the secant.

Answer 31

would it be 1/sqrt1-(x-1)^2???

Answer 32

if you meant to say
\[\Large \frac{1}{\sqrt{1-(x-1)^2}}\]
then you are correct

Answer 33

but my teacher does not like square roots as the denominator.

Answer 34

@jim_thompson5910

Answer 35

then multiply top and bottom by \(\Large \sqrt{1-(x-1)^2}\) to rationalize the denominator

Answer 36

doing that means
\[\Large \frac{1}{\sqrt{1-(x-1)^2}}=\frac{\sqrt{1-(x-1)^2}}{1-(x-1)^2}\]

Answer 37

All right. Thank you so much. I appreciate you taking the time to help me out!

Answer 38

no problem

Answer 39

WAIT

Answer 40

Can you tell me what x+2)^2+B^2=1 is?

Answer 41

i got sqrt1-(x-2)^2

Answer 42

solving \[\Large (x+2)^2+B^2=1\] for B gives \[\Large B = \sqrt{1-(x+2)^2}\]

Answer 43

all right thank you again!