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1. A horizontal plane is a plane parallel to the horizon.

2. A vertical plane is a plane perpendicular to a horizontal plane.

3. A horizontal line is a line parallel to a horizontal plane.

4. A vertical line is a line perpendicular to a horizontal plane.

5. A horizontal angle is an angle whose plane is horizontal.

6. A vertical angle is an angle whose plane is

vertical.

C

7. An angle of elevation is a verticle angle, one of whose sides is horizontal, and the inclined side above the horizontal side. Thus, BAC.

D

A

8. An angle of depression is a vertical angle, one of whose sides is horizontal, and the inclined side below the horizontal side. Thus, DCA.

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80. Problems.

1. Wishing to know the height of a tree standing on a horizontal plane, I measured from the tree the horizontal line BA, 150 ft., and found the angle of elevation, BAC, to the top of the tree to be 35° 20′. Required the height of the tree.

Ans. 106.335 ft.

B

2. In surveying a tract of land, I found it impracticable to measure the side AB on account of thick brushwood lying between A and B. I therefore measured AE, 7.50 ch., and EB, 8.70 ch., and found the angle AEB 38° 46'. Required AB.

Ans. 5.494 ch.

E

A

B

B

D

3. One side of a triangular field is double another, their included angle is 60°, and the third side is 15 ch. Required the longest side.

Ans. 17.32 ch.

4. Wishing to know the width of a river, from the point A on one bank to the point C on the other bank, I measure the distance AB, 75 yd., and find the angle BAC= = 87° 28′ 30′′, and the angle ABC= 47° 38' 25". Required AC, the width of

the river. Ans. 78.53 yd.

5. I find the angle of elevation, BAC, from the foot of a hill to the top to be 46° 25′ 30′′. Measuring back from the hill, AD-500 ft.,

I find the angle of elevation
ADC=25° 38' 40'. Required
BC, the vertical height of the
hill. Ans. 441.87 ft.

D

E

B

A

6. From the foot of a tower standing at the top of a declivity, I measured AB

45 ft., and the angle ABD 50° 15'. I also measured, in a straight line with AB, BC 68 ft., and the angle BCD = 30° 45'. Required AD, the

.height of the tower. Ans. 82.94 ft.

7. Wishing to know the height of a tower standing

on a hill, I find the angle of elevation, BAC, to the top of the hill to be 44° 35', and the angle of elevation to the top of the tower to be 59° 48'.

B

D

C

B

B

D

Measuring the

horizontal line AE, 275 ft., I find the angle of eleva

tion to the top of the tower to be 46° 25'. Required

the height of the tower.

Ans. 317.143 ft.

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33° 45', BDC 22° 30′. Re

Ans. DA=710.15 yd., DC=1042.5 yd., DB 934.28 yd.

1. a, NT

CM = cos a, covers a, OR cot a, CR sec a, CS

NO'

D

Remark. Describing the circumference through A, B, D, and drawing AE and BE, EAB BDC, EBA ADC.

Let a = the angle OCT
Then, we have MT

600 yd., BC=400 yd.,

=

RELATIONS OF CIRCULAR FUNCTIONS.

81. Fundamental Formulas.

MO

tan

=cosec a.

By articles 39-46, sin (90° — a) =

cos a, cos (90° — a) = sin a, etc.

From the diagram we have

D

= vers a,
O'S =

=

a,

E

the arc OT, and CO CT CN= sin

R

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с

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S

B

T

M

B

MT2 + CM2 = CT2.

Substituting the values of MT, CM, and CT, we have (1) sin2 a + cos2 a = 1.

Hence, The square of the sine of any arc plus the square of its co-sine is equal to 1.

From (1) we have, by transposition,

(2) sin2 a

(3)

Hence,

1. The square of the sine of any arc is equal to 1 minus the square of its co-sine.

cos2 a=

1— cos2 a,

1-sin2 a.

.'.

2. The square of the co-sine of any arc is equal to 1 minus the square of its sine.

From the diagram we have

MOCO- CM.

Substituting the values of MO, CO, and CM, we have

(4) vers a 1

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vers (90° — a)

(5) Covers α =

Hence, The versed-sine of any arc is equal to 1 minus its co-sine.

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COS a.

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1- sin a.

Hence, The co-versed-sine of any arc is equal to 1 minus its sine.

From the diagram we have

CM : CO:: MT : OR,

or cos a: 1 :: sin a tan a.
sin a

(6) tan a=

cos a

Hence, The tangent of any arc is equal to its sine

divided by its co-sine.

S. N. 7.

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