 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Inconvenient points (Posted on 2021-08-23) Two vertical poles stand 8.4m apart. AA' is 4.4m high, BB' is 3.1m high (A' and B' lying on the ground). A point P on the ground is defined to be a “convenient point”, if the viewing angle of points A and B from P is an acute one. If you move away from the poles, you can certainly find convenient points. There is a region of points on the ground, however, where all points are inconvenient.

Find the area of this region of inconvenient points.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) soln Comment 2 of 2 | We put pole base A' at the origin and pole base B' at b=8.4m on the x-axis.
The border of the inconvenient region is the locus of points P where the opening angle theta of vectors V1=(P-A) and V2=(P-B) is 90 degrees. Within this border the angle becomes obtuse.

V1 = (x, y, 0) - (0, 0, z1) = (x  , y, -z1)         z1 = 4.4m
V2 = (x, y, 0) - (b, 0, z2) = (x-b, y, -z2)        z2 = 3.1m

cos(theta) = V1 dot V2 / (mag V1) (mag V2)
cos(90) = 0 = V1 dot V2

x^2 - bx + y^2 + z1 z2 = 0

(x-b/2)^2 - (b/2)^2 + y^2 + z1 z2 = 0

(x-b/2)^2 + y^2 = (b/2)^2 -z1 z2
which is a circle of radius sqrt[ (b/2)^2- z1 z2] = 2m
centered at b/2 = 4.2m
(x-4.2)^2 - y^2 = 4
Area = 4 pi m^2

Edited on August 23, 2021, 11:21 pm
 Posted by Steven Lord on 2021-08-23 12:18:20 Please log in:

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