Although you haven't expressed it too clearly in your question, I think what you may be
looking for is the angle whose tangent is 4,900, and the angle whose tangent is 19,600 .
The problem is that the tangent of 89 degrees is about 57.3, and every number
from 57.3 all the way up to 'infinity' is the tangent of an angle somewhere in that
last degree, between 89 and 90 . . . They really bunch up in there !
4,900 is the tangent of 89.9883 degrees. (rounded)
19,600 is the tangent of 89.9971 degrees. (rounded)
tan-1(MUs)= critical angle
Take the inverse tangent -- tan-1(opposite side/adjacent side)
1.00
If you draw this angle in a coordinate system, a right triangle is formed in the second quadrant, with length legs 12 and 5 units. If you label with O the angle that is formed by the terminal side and y-axis, you have tan O = 12/5 and O = tan-1 12/5 = 67.38 degrees Thus, the given angle has a measure of 157.38 degrees (90 + 67.38).
It is: tan^-1*(0.8) = 38.7 degrees
sin, tan and cos can be defined as functions of an angle. But they are not functions of a triangle - whether it is a right angled triangle or not.
I can't either. Don't know what a "tan angle" is, so don't know what I'm looking for. It might be right here under my nose.
tan u/2 = sin u/1+cos u
Using trigonometry its height works out as 63 meters to the nearest meter. -------------------------------------------------------------------------------------------------------- let: h = height building α, β be the angles of elevation (29° and 37° in some order) d be the distance between the elevations (30 m). x = distance from building where the elevation of angle α is measured. Then: angle α is an exterior angle to the triangle which contains the position from which angle α is measured, the position from which angle β is measured and the point of the top of the building. Thus angle α = angle β + angle at top of building of this triangle → angle α > angle β as the angle at the top of the building is > 0 → α = 37°, β = 29° Using the tangent trigonometric ratio we can form two equations, one with angle α, one with angle β: tan α = h/x → x = h/tan α tan β = h/(x + d) → x = h/tan β - d → h/tan α = h/tan β - d → h/tan β - 1/tan α = d → h(1/tan β - 1/tan α) = d → h(tan α - tan β)/(tan α tan β) = d → h = (d tan α tan β)/(tan α - tan β) We can now substitute the values of α, β and x in and find the height: h = (30 m × tan 37° × tan 29°)/(tan 37° - tan 29°) ≈ 63 m
tan-1(MUs)= critical angle
It can be shown that:height = (d tan α tan β)/(tan α - tan β)where: α is the angle closest to the objectβ is the angle further away from the objectd is the distance from the point of angle α to the point of angle βThus: height = (80 ft × tan 45° × tan 34°)/(tan 45° - tan 34°) ≈ 165.78 ft
It is a trigonometric equation for a right triangle, to find a non-right-angle angle. Using SOHCAHTOA, it is the opposite side divided by the adjacent angle
tan-1(0.4877) = 25.99849161 or about 26 degrees
It can be shown that:height = (d tan α tan β)/(tan α - tan β)where: α is the angle closest to the objectβ is the angle further away from the objectd is the distance from the point of angle α to the point of angle βThus: height = (53 ft × tan 31.4° × tan 26.4°)/(tan 31.4° - tan 26.4°) ≈ 140.87 ft
It can be shown that:height = (d tan α tan β)/(tan α - tan β)where: α is the angle closest to the objectβ is the angle further away from the objectd is the distance from the point of angle α to the point of angle βThus: height = (40 ft × tan 50° × tan 30°)/(tan 50° - tan 30°) ≈ 44.80 ft
If you know the gradient for a line (the m in y = mx + c) then tan-1 (m) will give you the angle between the line and the x axis. So do tan-1 for both gradients and subtract to find angle between the lines.
You have to use trig. If the base angle is a and base b, the height is b tan(a).