Small angle (1 Viewer)

[atakach99]

maybe i should post this question in another thread u guys seem busy

 

[tommykins]

ok this one is seems easy but everytime i try it i get a different answer from the back of the book

Find the diameter of the sun to the nearest kilometre if its distance from the earth is 149000000 km and it subtends an angle of 31 ' at the earth

plz help
thnkss

It draws out a right angled triangle with 149 000 000 km as the hypotenuse. The diameter of the sun is the "x" (opposite side) and the 31' is the angle.

Convert 31' into radians.

Since sin x = Opposite/Hypotenuse, then sin 31' = x/149 000 000

thus x = 149 000 000 * sin 31'

I somehow need to get rid of the bottom one (or change it) so that I can work it out. It's not as easy as it looks like. I bloody used the t-formula and it gets more complicated.....

If it's too hard, i'll post up solution.

 

[lyounamu]

If it's too hard, i'll post up solution.

Thanks. :uhhuh:

I would love the solution as it will relieve the pain that dwells in my brain.

 

[tommykins]

Thanks. :uhhuh:

I would love the solution as it will relieve the pain that dwells in my brain.

lim x -> pi/4 sinx - cosx / x - pi/4

sinx - cosx = sqrt2 sin(x-pi/4) <--- solutions lost me from here.

lim x -> pi/4 sinx - cosx / x - pi/4 = lim x -> pi/4 sqrt2 sin(x-pi/4)/x - pi/4 = sqrt2 * 1 = sqrt2.

Once again, don't know how on earth it works.

 

[lyounamu]

lim x -> pi/4 sinx - cosx / x - pi/4

sinx - cosx = sqrt2 sin(x-pi/4) <--- solutions lost me from here.

lim x -> pi/4 sinx - cosx / x - pi/4 = lim x -> pi/4 sqrt2 sin(x-pi/4)/x - pi/4 = sqrt2 * 1 = sqrt2.

Once again, don't know how on earth it works.

I get it now. Thanks!!

sinx - cosx = square root 2 . sin(x)cos(pi/4) - square root 2 . cos(x)sin(pi/4))
= square root 2 . (sin(x)cos(pi/4) - cos(x)sin(pi/4))
= square root 2 . sin (x-pi/4)

Then lim x - > pi/4 (sin - cos x)/ (x-pi/4) = (square root 2 . sin(x-pi/4)) / (x - pi/4)
= square root 2 . 1 = square root 2