Question

please help????? i attached a picture. i need someone to explain a little of it

Answer 1

Use SOH CAH TOA? do you understand this?

Answer 2

noo:(

Answer 3

anyone?

Answer 4

do you know trignometry????

Answer 5

how 'bout using similarity with the 45-45-90 triangle whose sides are \(\large \frac{\sqrt2}{2} \), \(\large \frac{\sqrt2}{2} \), 1

Answer 6

i know how to use sin tan cos with some problems but i keep messing up on this one

Answer 7

let unknown side be x......
cos 45=x/23
\[\frac{1}{\sqrt{2}}=\frac{x}{23}\]
\[x=\frac{23}{\sqrt{2}}\]

Answer 8

or if angles are in the ratio 1:1:2 ....sides are in the ratio
\[1:1:\sqrt{2}\]

Answer 9

ok here is the solution what you have is a right angle isosceles triangle so angle ACB is also 45 degrees
now
BC=23 ft= hypotneuse
Angle ACB= 45
You have to find the height

Use sine here
Since \[\sin = height \div hypotenuse \]
\[ \sin 45= 23\div h (where h=hyptenuse)\]

\[Sin 45 =1\div \sqrt{2}\]

Answer 10

now (1/sqrt(2))= h/23
h=23/sqrt(2)

Answer 11

@The_Prophet..will you tell me where you got the 1/\_2

Answer 12

@Aria11 there is a table of trigonometric standard angles google it

Answer 13

cos 45=1/sqrt 2.......by definition .....u need to learn it...

Answer 14

@Aria11 from there @THE_PROPHET and I agree with prophet you need to learn it

Answer 15

ok ty imp1988 and the_prophet