Computing Angles of Rotation and Angles of Tilt In each of the following problems, the axis of a hole is shown in a rectangular solid. In order to position the hole axis for drilling, the angle of rotation and the angle of tilt must be determined. Compute angles to the nearer minute in triangles with customary unit sides. Compute angles to the nearer hundredth degree in triangles with metric unit sides. a. Compute the angle of rotation, R. b. Compute the angle of tilt, T. 7. Given: H= 2.600 in. L = 2.400 in. a. W= 1.900 in. 8. Given: H= 55.00 mm b. Use this figure for #7 and #8. AXIS OF HOLE L 48.00 mm W= 30.00 mm H a. b. 9. Given: H = 4.750 in. L = 4.000 in. W= 3.750 in. a. 10. Given: H=42.00 mm b. L37.00 mm W = 32.00 mm a. b. 11. Given: H = 0.970 in. L = 0.860 in. W= 0.750 in. a. 12. Given: H= 22.00 mm L 18.00 mm = W = 15.00 mm a. b. Use this figure for #9 and #10. ZR AXIS OF HOLE Use this figure for #11 and #12. H b. L AXIS OF HOLE T Next Unit md 2 April Tan (

Welding: Principles and Applications (MindTap Course List)
8th Edition
ISBN:9781305494695
Author:Larry Jeffus
Publisher:Larry Jeffus
Chapter20: Shop Math And Weld Cost
Section: Chapter Questions
Problem 12R: Find the area of the following: a. Square that is 55 wide b. Circle with a 22 diameter c....
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Just do Questions 7, 9, 11. Here are notes attached for reference. I prefer handwritten solutions. ONLY UPLOAD A SOLUTION IF YOU ARE SURE ABOUT THE ANSWER PLEASE.

Computing Angles of Rotation and Angles of Tilt
In each of the following problems, the axis of a hole is shown in a rectangular solid. In
order to position the hole axis for drilling, the angle of rotation and the angle of tilt must be
determined. Compute angles to the nearer minute in triangles with customary unit sides.
Compute angles to the nearer hundredth degree in triangles with metric unit sides.
a. Compute the angle of rotation, R.
b. Compute the angle of tilt,
T.
7. Given: H= 2.600 in.
L = 2.400 in.
a.
W= 1.900 in.
8. Given: H= 55.00 mm
b.
Use this figure for #7 and #8.
AXIS OF HOLE
L 48.00 mm
W= 30.00 mm
H
a.
b.
9. Given: H = 4.750 in.
L = 4.000 in.
W= 3.750 in.
a.
10. Given: H=42.00 mm
b.
L37.00 mm
W = 32.00 mm
a.
b.
11. Given: H = 0.970 in.
L = 0.860 in.
W= 0.750 in.
a.
12. Given: H= 22.00 mm
L 18.00 mm
=
W = 15.00 mm
a.
b.
Use this figure for #9 and #10.
ZR
AXIS OF HOLE
Use this figure for #11 and #12.
H
b.
L
AXIS OF HOLE
T
Transcribed Image Text:Computing Angles of Rotation and Angles of Tilt In each of the following problems, the axis of a hole is shown in a rectangular solid. In order to position the hole axis for drilling, the angle of rotation and the angle of tilt must be determined. Compute angles to the nearer minute in triangles with customary unit sides. Compute angles to the nearer hundredth degree in triangles with metric unit sides. a. Compute the angle of rotation, R. b. Compute the angle of tilt, T. 7. Given: H= 2.600 in. L = 2.400 in. a. W= 1.900 in. 8. Given: H= 55.00 mm b. Use this figure for #7 and #8. AXIS OF HOLE L 48.00 mm W= 30.00 mm H a. b. 9. Given: H = 4.750 in. L = 4.000 in. W= 3.750 in. a. 10. Given: H=42.00 mm b. L37.00 mm W = 32.00 mm a. b. 11. Given: H = 0.970 in. L = 0.860 in. W= 0.750 in. a. 12. Given: H= 22.00 mm L 18.00 mm = W = 15.00 mm a. b. Use this figure for #9 and #10. ZR AXIS OF HOLE Use this figure for #11 and #12. H b. L AXIS OF HOLE T
Next Unit
md
2 April
Tan (<CAB) =
(8)
Tan (CCAB) =
opp
opp
まる
adj
Compute Diagonal Lengths and Angle Sizes in
Rectangular Solids
5
50
२०
2
Tan ( 0 ) = 2.5
=Tan (2.5)
<CAB = 68...
10
111·4.02
90270
ABAC²+CB
AB² = 15² + 114.02²
AB²=5625+12446
JAB U18621
8100
Tan (c)= opp
adj
Tan (CR) = 80
Too
= 80²+10
| AB = 50 41 28.05²
2
=6400+ 1000co AB = 2500+1894.36
=
16400
=V16900
= 128.06 cm
AB = 13747
Tan(CT)=12106
"Tan (CR) = 0.8 109 1103
CR Tan Co.
CR = 38.65
50
CT-68.67%
70
CB-90²+702
Co²-11074900
90
CB²=13800
CB= 114.02
Tam (CA)=70
५०
90
CR-37.17
A
13646B
tan (<T) = 114.02
114.02
Tan (CA)=70
<R=37.87°
CT = 56.65°
AB=136.46
Transcribed Image Text:Next Unit md 2 April Tan (<CAB) = (8) Tan (CCAB) = opp opp まる adj Compute Diagonal Lengths and Angle Sizes in Rectangular Solids 5 50 २० 2 Tan ( 0 ) = 2.5 =Tan (2.5) <CAB = 68... 10 111·4.02 90270 ABAC²+CB AB² = 15² + 114.02² AB²=5625+12446 JAB U18621 8100 Tan (c)= opp adj Tan (CR) = 80 Too = 80²+10 | AB = 50 41 28.05² 2 =6400+ 1000co AB = 2500+1894.36 = 16400 =V16900 = 128.06 cm AB = 13747 Tan(CT)=12106 "Tan (CR) = 0.8 109 1103 CR Tan Co. CR = 38.65 50 CT-68.67% 70 CB-90²+702 Co²-11074900 90 CB²=13800 CB= 114.02 Tam (CA)=70 ५० 90 CR-37.17 A 13646B tan (<T) = 114.02 114.02 Tan (CA)=70 <R=37.87° CT = 56.65° AB=136.46
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