A new satellite network tower was installed in the non receptive areas of mabatang and calaguiman.the satellite in branched by two wires fastened to the ground at the same point.the shorter wire is 30 ft long and the fastened to the tower 10 feet above the foot of the tower.the longer one is 33 ft long and is fastened to the tower 17 ft above the foot of the pole. Find the exact value of sin x-y, the sine angle between two wires.and find CSC x and sec y?
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Please answer my problem question
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I have 2 right-angled triangles, with a common base of k ft
I see two sine values:
sin x = 17/33
sin y = 10/30
using the smaller triangle: k^2 + 10^2 = 30^2
k = √800 = 20√2
using the larger triangle: k^2 + 17^2 = 33^2
k = √800 = 20√2 , well that is good, they are the same.
so cosx = 20√2/33
cosy = 20√2/30
sin(x-y) = sinx cosy - cosx siny
= (17/33)(20√2/30) - (20√2/33)(10/30)
= 34√2/99 - 20√2/99
= (14√2)/99
cscx = 1/sinx = 1/(17/33) = 33/17
secy = 1/cosy = 30 / (20√2)
I have 2 right-angled triangles, with a common base of k ft
I see two sine values:
sin x = 17/33
sin y = 10/30
using the smaller triangle: k^2 + 10^2 = 30^2
k = √800 = 20√2
using the larger triangle: k^2 + 17^2 = 33^2
k = √800 = 20√2 , well that is good, they are the same.
so cosx = 20√2/33
cosy = 20√2/30
sin(x-y) = sinx cosy - cosx siny
= (17/33)(20√2/30) - (20√2/33)(10/30)
= 34√2/99 - 20√2/99
= (14√2)/99
cscx = 1/sinx = 1/(17/33) = 33/17
secy = 1/cosy = 30 / (20√2)
sinx = 17/33, so cosx = √800/30
siny = 10/30, so cosy = √800/33
Now just use your trig formulas.
sin(x-y) = sinx cosy - cosx siny
cscx = 1/sinx
secy = 1/cosy
siny = 10/30, so cosy = √800/33
Now just use your trig formulas.
sin(x-y) = sinx cosy - cosx siny
cscx = 1/sinx
secy = 1/cosy
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