By Carl E. Smith

Arithmetic, advisor, How-to, Communications, Engineering

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If any two sides of a right triangle are given the angles can be found by using one of the above equations and then looking in a table of trigonometric functions for the angle. hyp = o~p sm opp = sin hyp adj cos = --hyp adj hyp = cos adj = cos hyp tan = ()~~ adJ d' opp aJ=tan opp = adj tan (8) (9) (10) (ll) (12) Example 1. In Fig. 25 we have given adjacent side a = 8 hypotenuse c = 10 Find opposite side b angle a (alpha) angle fJ (beta) Solution. By Eq. 8 hyp c 10 Adjacent side for {3 b Opposite side fora 90° a (14) (15) a Adjacent side for a Opposite side for {3 FIG.

B. 32 = A. 737 l' B:tJ The sum of the y-components by Eq. (2) is AV\~:tJ X R;/O FIG. -Vector diagram Example 2. for These components are shown in Fig. 30. The magnitude of the resultant vector R by Eq. (3) is Find the vector sum of A = 8/20° and B = 20/120°. Solution. 057)2 and y-components; thus, by Eq. 5175 By Eq. 737 It will be noted that vector B is FIG. -Vector diagram for Ex- in the second quadrant, so the x-com­ ample 2. ponent will be along the negative xaxis. In this case we subtract the angle from 180° and get 180° - 120° = 60° This is the angle that vector B makes with the negative x-axis.

In words this says that the resultant magnitude y is equal to the square root of the sum of the squares of the magni­ P tudes of the two vectors. See Fig. 11 for a similar illustration. -When the _I"< ! vectors are not at right angles, as F IG. 26 . -Vect or f orees P and Q illustrated in Fig. 26, it is customacting on the point 0 at angles of ary to resolve the vectors into a and {j respectively. their components along the x-axis and the y-axis. The algebraic sum of the x-components gives a resultant x-component and the algebraic sum of the y-component gives a resultant y-component.

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