1. − 165 ° + 360 ° − 165 ° − 360 ° = = 195 ° − 525 °
2. 39 ′ = (3960) ° = 0.65 ° 37 ° 39 ′ = 37.65 °
3. tan 73.8 ° = 3.44
4. cos θ = 0.3726 ; θ = 68.12 °
5. Let xx22.62x = = = = distance from course to eastsin 4.05 ° 22.62 sin 4.05 ° 1.598 km
6. sin θ = 23x = 32 − 22−−−−−−√ = 5–√tan θ = 25–√
7. tan θ = 1.294 ; csc θ = 1.264
8. Ba52.8a = = = = 90 ° − 37.4 ° = 52.6 ° tan 37.4 ° 52.8 tan 37.4 ° 40.452.8cc = = = cos 37.4 ° 52.8cos 37.4 ° 66.5
9. 2.492 + b2 = 3.882b = 3.882 − 2.492−−−−−−−−−−−√ = 2.98sin A = 2.493.88A = sin − 1(2.493.88)A = 39.9 ° B = 50.1 °
10. s / 212.0 = cos 42.0 ° s = 24.0 cos 42.0 ° = 17.8
11.
h = 92 + 402−−−−−−−√ = 41sin θ = 941 , cos θ = 4041sin θcos θ = 9/4140/41 = 940
12. λ = = = d sin θ30.05 sin 1.167 ° 0.6120 μm
13. r = 52 + 22−−−−−−√ = 29−−√sin θ = 229−−√ = 0.3714cos θ = 529−−√ = 0.9285tan θ = 25 = 0.4000csc θ = 29−−√2 = 2.693sec θ = 29−−√5 = 1.077cot θ = 52 = 2.500
14. Let x = new length of ramp
sin 4.50 ° = 2.50xx = 2.50sin 4.50 ° = 31.9 ftadded length = 31.9 − 9.5 = 22.4 ft
15. Distance between points is x − y .
18.525x = tan 13.500 ° 18.525y = tan 21.375 ° x − y = 18.525tan 13.500 ° − 18.525tan 21.375 ° = 29.831 m
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