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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.
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.
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.
1. Wishing to know the height of a tree standing on a horizontal plane, I meas
ured 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.
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.
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
6. From the foot of a tower standing at the top of a declivity, I measured AB
.height of the tower. Ans. 82.94 ft.
7. Wishing to know the height of a tower standing
horizontal line AE, 275 ft., I find the angle of eleva
tion to the top of the tower to be 46° 25'. Required Ans. 317.143 ft.
the height of the tower.
ADC 33° 45', BDC22° 30′. Required DA, DC, DB.
Ans. DA=710.15 yd., DC=1042.5 yd., DB 934.28 yd.
Remark. Describing the circumference through A, B, D, and drawing AE and BE, EAB BDC, EBA ADC.
cot a, CR sec a, CS: =cosec a.
By articles 39-46, sin (90°— a) =
cos a, cos (90° — a) — sin a, etc.
From the diagram we have
MT2 + CM2 = CT2.
Substituting the values of MT, CM, and CT, we have
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,
1. The square of the sine of any arc is equal to 1 minus the square of its co-sine.
2. The square of the co-sine of any arc is equal to 1 minus the square of its sine.
Substituting the values of MO, CO, and CM, we have
Hence, The versed-sine of any arc is equal to 1 minus its co-sine.
Hence, The co-versed-sine of any arc is equal to 1 minus its sine.
From the diagram we have
Hence, The tangent of any arc is equal to its sine
divided by its co-sine.
S. N. 7.