Question

I have 3 d trigonometric problem with a drawing. How can I attach the file containing the image of the problem. Any one please guide me

Answer 1

use the attach file option ....

Answer 2

Thank you so much. I am attaching the file containing the problem. I am unable to find any solution to this. Kindly help me.

Answer 3

the person is always 63m high, the distance of the base is always changing .... which gives us a changing tangent angle of elevation:

tan(R) = 63/B, right?

Answer 4

defining the base measure in ... some terms, sems beneficial to me

Answer 5

derivatives help to define min/max points

tan(R) = 63/base

base tan(R) = 63

base' tan(R) + R' base sec^2(R) = 0

R' = -(b' tan(R))/(b sec^2(R))

that might play a part

Answer 6

i am getting clueless

Answer 7

i started clueless :)

i determine that

KR = 58.31
RN = 70.71
KN is given already as 80 for this base triangle setup right?

Answer 8

yeah

Answer 9

defining the angle of depression at R is a matter of getting a setup using this data

Answer 10

the 80-d is a bad thought, just d for the distance walked, 80-d is the other side if we need to use it. which might be easier since its got a 45 .... lets try it

Answer 11

thats inspirational

Answer 12

using law of cosines:\[b^2=(80-d)^2+(70.71)^2-2(80-d)(70.71)\cos(45)\]
to define the base part with for some given distance walked

Answer 13

great going

Answer 14

and the angle at R is equal to the arctangent of 63/b
\[R=tan^{-1}\left(\frac{63}{\sqrt{(80-d)^2+(70.71)^2-\sqrt2(80-d)(70.71)}}\right)\]
looks fun

Answer 15

this is a real tough nut to crack

Answer 16

another idea: the angle of depression get bigger the smaller we get the base


so it really boils down to our definition of b in terms of d