Maths URGENT HELP Plz

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For this one, I need only to know the (a) (iii) the shortest distance from C to BD

& part (b) The greatest angle of depression..

Plzzzzz
 

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bamteck said:
Plz Plz Help ::
What difficulty do u have??
(i) 6
(ii) (a)Area of circle is nr^2....the diameter of the largecircles is equal to the 3 diameters of small circle .....hence find the radius...and then find the area!!
(b) find the total area of small circles ...find the difference b/w large circle area and total area of small ones!!this gives the left out part...if u notice there are 6 such regions similar to shaded one so if u divide the difference by 6 u will get the shaded area :)
i think u might be finding this part difficult!! hope u r clear with it now...i did nt give u the answer cuz u shud try it on ur own!!I can help u but its always better to it urself :)
 

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bamteck said:
For this one, I need only to know the (a) (iii) the shortest distance from C to BD

& part (b) The greatest angle of depression..

Plzzzzz
Remember...the shortest distannce is always the perpendicular distance!!
so then u can use the trignometric ratios....in this case sin0 = opp/hyp...and find opp when hyp = 1040 and 0= 42 degrees!!
 
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Rightly said math angel! As for the next question bamteck,
the shortest distance from C to BD is the perp. distance from C to BD. Since u knowangle CBD and BC, the answer would be: sin 42= x/1040.
The greatest angle of depression would occur where the perpendicular distance from C to BD is shortest. So tan x =500/perp. distance found in (a) (iii).
Hope this helps :)
 

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I suppose the question to be done in this way:
vol.of hollow cylinder = (pi) x h x (R^2 - r^2)
Where R is the radius of big circle and r is the radius of small circle.....so R = r + 1.5
since we know vol. as 500...R=1.5 + r .....h = 6 solve the equation!!

500 = (pi) x 6 x [(1.5 + r)^2 - r^2]

500/[6(pi)] = r^2 + 3r + 2.25 - r^2
500/[6(pi)] = 3r + 2.25

r = 8.0919 = 8.09

Note...(r + 1.5 ) ^2 = r^2 + 3r + 2.25 ....[recall expanding (a + b)^2




Actually i havent understand what Uxair has done...but i'm quite sure my answer and method is correct!!
 

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it's quite easy...for c u have to plot the points given in the grid above...for d just draw a tangent to the curve at the given point....and find the gradient of the line......for e (i) draw the line when the eqn of line is what is given....(by plotting the grid)
(ii) then write the point where te line intersects the graph....
(iii) equate both the eqns. ....the eqn of curve = 1/2 x + 3
and simpligy to make it in the form required...
hope u understood :)
 
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