|
p
<br />r
<br />17
<br />21
<br />_/42
<br />- - --
<br />I
<br />TRICONOMETRIC FORMULA; :.
<br />B B B
<br />�t
<br />i a a c a c
<br />a
<br />b .� A
<br />C
<br />L � Rigbt Triangle - Oblique Triangles
<br />Solution of Right Triangles
<br />b a b c
<br />For Angle Aa, sin — � ,cos= —,tan= T ,cat = a ,sec = �c, cosec = --
<br />Given Required a
<br />a,b A,B,c tanA==cotB,c= u� }b =rz li a2
<br />.I a,e A,B,& sinA=c=Cos B,b+a)(e—a)=a�1—os
<br />I' --
<br />A, a
<br />A,
<br />B, b, c
<br />B=90°—A, b= a cotA, c= a
<br />Area
<br />s— 2 , area = s(v
<br />sin A.
<br />A, b
<br />B, a, c
<br />B=90°--A,a = btanA,o= b b,
<br />bcsinA D
<br />area =
<br />/
<br />C6S A.
<br />A, a
<br />B, a, b
<br />B = 90°—A, a = c sin A, b = c cos A,
<br />�
<br />Solution
<br />of Oblique Triangles
<br />Given
<br />B' a
<br />Required
<br />b, c' C
<br />_ a sin B a sin C
<br />b
<br />' C = 180°—(A f B), c =
<br />sin A sin A
<br />A, a, b
<br />B, c, C
<br />b sin A a sin C
<br />sin B =
<br />,C = 180°---{A + B), c =t.
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the verticalanale.Tbus: slope distance =3o9.4fL
<br />1 ,:• e
<br />+-we. ? �A
<br />awe.,
<br />a sin A,
<br />a, b, C
<br />A, B, c
<br />A+B=180°— C, tan a (A—B)= (a—b) tan �, (A+B)
<br />¢+b
<br />a sin 0
<br />slit ft
<br />a, b, a
<br />A, B, C
<br />s— ,sin aA--\I ,
<br />2 b o
<br />k a 8—i
<br />sin'B—.���` ��c i,C-1$0°�A-}•B)
<br />A. distance times (l.—cosmn of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained- Cosine 5°10'=.9959.1—.99:19=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=$18.09 ft. r.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope dis#anee=302.6 ft. Horizontal distance=30214X14=
<br />6— =302.6-0.M=30128 ft.
<br />3 X 302.6 + .
<br />' '' 61A6E IN Ef. B. A. ,i.•i
<br />a, b, c
<br />Area
<br />s— 2 , area = s(v
<br />f
<br />A, b, e
<br />Area
<br />bcsinA D
<br />area =
<br />/
<br />2
<br />�
<br />A., B, C, a
<br />Area
<br />' ¢'- sin B sin C
<br />area — 2 sin d
<br />___-_ -------_---.._.._.�.,�,�.•,_;:�I"
<br />RRDUCTION TO HORIZONTAL
<br />`
<br />�6
<br />Horizontal distance= Slope distance multiplied by the
<br />cosine of the verticalanale.Tbus: slope distance =3o9.4fL
<br />1 ,:• e
<br />+-we. ? �A
<br />awe.,
<br />SE�p
<br />� 5104 ��g�e
<br />Vert. angle =51 19. From Table, Page IX. cos 5'1(Y=
<br />9959. Horizontal distance=319.4X.9959=318.09 ft.
<br />� Horizontal distance also=Slope distance minus slope
<br />A. distance times (l.—cosmn of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained- Cosine 5°10'=.9959.1—.99:19=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=$18.09 ft. r.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft.,
<br />slope dis#anee=302.6 ft. Horizontal distance=30214X14=
<br />6— =302.6-0.M=30128 ft.
<br />3 X 302.6 + .
<br />' '' 61A6E IN Ef. B. A. ,i.•i
<br />
|