CPhill

Of my avatar or of the lady looking at the flagple??

Note........they could be one and the same.......

I see a way that I could have made this easier........let's go from this step......

tan(14) = x/d   ........let's use this instead........x = d*tan(14)

And in the next part I had

tan(18) = (40-x) / d             let's substitute  d*tan(14) for "x'

So we have.....

tan(18) = (40- d*tan(14)) /d     ...multiply both sides by d

d*tan(18) = 40 - d*tan(14)       .....add d*tan(14) to both sides

d*tan(18) + d*tan(14) = 40   ....factor the "d" out on the left side

d[tan(18) + tan(14)] = 40         ..........divide both sides by [tan(18) + tan(14)]

d = 40 / [tan(18) + tan(14)] = 69.7 ft

This way....we don't have to worry with "x" at all !!!

Anonymous...this problem is quite a bit more difficult than the previous ones you have submitted.....I'll explain it...if you have trouble...let me know...it requires some "substitutions"

First.....as the woman looks at the pole...let's call the the opposite side to the angle of depression, "x"......this "x" represents only a part of the pole height that is below her sight line....so we have

tan(14) = x/d   where "d" is the distance the woman is from the pole ...this means that  

d = x/tan(14)

Now.........the part of the pole that is above her sight line is just 40-x...so using the angle of elevation and the tangent again, we have

tan(18) = (40-x)/d

Now....because  I want an equation in one variable, I'm going to substitute x/tan(14) for "d"in this second equation...so we have

tan(18) = (40-x)/ (x / tan(14))      ...  multiply both sides by (x/tan(14))..this gives

tan(18) (x / tan(14) = (40-x)         .... doing a little "rearranging" we have

[tan (18) / tan (14)] x   = 40- x      ......divide tan(18) by tan(14) = 1.303

1.303x = 40 -x                              .......add x to both sides

2.303x = 40                                  ........divide both sides by 2.303

x = 40/2.303 = 17.37                     this "x" represents the height of the pole below her sight line........but that's not what we're after...we want "d"...so using d = x/tan(14), we have

d = 17.37 / tan(14) = 69.7 ft.......and that's her distance from the pole

I realize you might get "stuck" on this one.......let me know....

OK...assuming the frame width is uniform..let's call this "x"

So....the total area that the frame encloses is 18 * 14.

So......the area occupied by the picture itself can be expressed as

(18-2x)(14-2x) = 165       

252 - 64x + 4x^2 = 165    subtract 165 from both sides

4x^2 - 64x + 87 = 0          factor this.....

(2x - 29) (2x -3) = 0   set each factor to 0

2x - 29 = 0      x =14.5 cm  reject this.........the frame width would cover the picture !!!

2x - 3 = 0        x = 1.5 cm......that seems reasonable

Also (18 - 2(1.5)) * ( 14 - 2(1.5))  =  (18 -3) * ( 14 -3) =  (15) * (11) = 165 cm

This seems to "sync-up" with the info given

So the frame is 1.5 cm wide.

Winner

Winner

Chicken

Dinner !!!!

"Thums Up" and Pernts from me, GL !!!

Angle BCD is also 30.3  So..BDC = 90-30.3 = 59.7

So we can find BC with the Law of Sines

BC/sin59.7 = 67.7 / sin 90

BC = 67.7* sin 59.7  = 58.5 cm

So AC - BC = AB =  67.7 - 58.5 = 9.2 cm

Notice that this doesn't seem to square with the picture as presented...the probable reason is that "theta" seems much greater than 30.3 degrees......However, we can only work with what we're given....

I couldn't have done it without some help from GL........

No...we already know the angle (that's what the "arc" functions tell us)

Just solve

tan(12.5) = 4 / x     

for "x"

Let's look at this....you're using the correct function.....note that the height of the basket doesnt change.....and we're looking for the adjacent side....note that this isn't 15 anymore, because the player has moved back from the FT line......

So....

tan 12.5 = 10/x

Note that x must now be greater than 15 .....can you take it from here??

Whoops...I made a mistake...I forgot that we said that the player's eyes are 6 ft from the floor....disregard the previous baloney.......

Note now that the basket is only 4ft feet above the player's line of sight..

So we have...

tan12.5 = 4/x

OK...I checked this answer...it makes sense........see if you can tell me what we get now for "x".......

Good job, Anonymous.......continue on !!