Trigonometric grade 11 math(uni) word problems
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Trigonometric grade 11 math(uni) word problems

[From: ] [author: ] [Date: 11-05-20] [Hit: ]
-This is a fairly easy question.You have to calculate the height of the end of the boom at 30 deg. elevation and again at 45 deg. and then subtract one from the other.The height at 30 deg. is given by the equation 12 sin 30= height which is 12 times 1/2 or 6 meters.......
Im having problems with these words problems i dont know why but my answers not coming out the way it should from what the back of the text book says.

The arm of a boom crane is 12m long because of the location of the construction site, the angle of inclination of the boom of the crane is has min. of 30 degrees and max of 45 degrees find the vertical displacment of the end of the boom as

a) an exact value
b) appproximate value

for this question i got y1-y2=6(1-(square root of 3) m
y2-y1=4.4m

however the correct answer is 6(square root of 2 -1) and 2.5m ?

-
This is a fairly easy question. You have to calculate the height of the end of the boom at 30 deg. elevation and again at 45 deg. and then subtract one from the other.

The height at 30 deg. is given by the equation 12 sin 30 = height which is 12 times 1/2 or 6 meters.

The height at 45 deg. is 12 sin 45 or 12 times (sqrt2)/2 which is 6 sqrt2 meters.

Now subtract... 6 sqrt2 - 6 = 6((sqrt2) - 1). When you plug in an actual value for sqrt 2 it comes out to a hair under 2.5 meters or about 2.48 meters.

-
1st note that 1 - √3 < 0 and so cannot be correct....

do you not know the values of the sine function ?? 6 ( √2 - 1 ) ≈ 2 .48
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