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4. The angle of elevation of the top B of a tower AB from a point X on the ground is 60°. at a point Y 40m vertically above X, the angle of elevation of the top is 45°. find the height of the tower AB and the distance XB.

Let D be the horizontal distance from X to the tower base

Let H be the tower height

And we have :

tan 60  = H/D  →  D = H / tan 60

And we know that....the vertical difference between the tower's height and Y = (H - 40)...so...

tan 45  = (H - 40) / D       sub for D

tan 45  = (H - 40) / [H/tan 60 ]

tan 45  = tan 60 *(H - 40) / H

H *tan 45  = tan 60 (H - 40)

H * tan 45  = H* tan 60  = 40 tan 60

H = H * sqrt(3) - 40sqrt(3)

40sqrt3 = H [ sqrt(3) - 1]

[40sqrt(3)] /[ sqrt(3) -1]  = H  ≈ 94.64 m

And XB ≈ H csc 60 ≈ 94.64 csc 60 ≈ 109.28 m

Here's a pic....

re-arrange    -6x^2 + 36x =-216       Divide by -6  (we'll multiply by -6 shortly to even things out)

                        x^2 -6x =  36          'complete the square'

                        x^2 - 6x + 9 = 36+9

                        (x-3)^2 -45    now multiply by -6

                        -6(x-3)^2 + 270       a = -6   b = -3  c = 270

                    a+b+c = 261

3. From the top of the tower the angle of depression of an object on the horizontal ground is found to be 60°. on decending 20m vertically downwards from the top of the tower, the angle of depression of the object is found to be 30°. find the height ofthe tower.

Let O   be the object's distance from the tower's base  and H  be the tower height....and we have that

tan (60)  = H/O   →  O  = H/tan (60) = H/sqrt(3)

And

tan (30)  = (H - 20) / O              sub  H/ tan(60) for O

tan (30)  = (H - 20) / [H / tan(60)]

tan (30)  = tan(60) ( H - 20) / H

H tan (30)  = tan(60) (H - 20)

Htan(30) = Htan(60) - 20tan(60)

20  tan(60)  = H tan(60) - H tan(30)

20tan(60)  = H [tan (60) - tan (30)]

[20tan(60) ]/ [ tan(60) - tan (30) ]  = H  = 30 m

Here's a pic  AB  = tower height  = 30m   ....OA  = H/sqrt(3)  = 30/sqrt(3)

55!/(51! x 4!)  = (55 x 54 x 53 x 52 x 51!) / (51! x 4!) = (55x54x53x52)/4! = 341055

  (as guest found...just thought I would show a little more of the reason/method) 

When a 15 newton force is applied to on end of a lever 100 cm long, a 60 newton weight is attached to the opposit end is lifted. What is the length of the input arm? {nl} Please show all the steps.

\(15N\times x=60N\times(100cm-x)\)

\(15N\times x = 6000Ncm - 60N\times x\)

\(x=\frac{6000Ncm}{75N}\)

\(x=80cm\)

\( The\ length \ of \ the\ input\ arm \ is \ 80cm.\)

  !

Say "I eat Apples on A pirate Ship" while holding your tongue

2. The horizontal distanc between two towers is 60m. the angle of elevation of the top of the taller tower as seen from the top of the shorter one is 30°.If the height of the taller tower is 150m, then find the height of the shorter tower.

The height of the shorter tower  =

150 - 60tan30  =  150 - 60sqrt(3)  ≈  115.36 m

Here's a pic....BA   = shorter tower = 115.35m  .....CD = taller tower = 150m....angle CAF = 30.....AF = 60m

Thanks so much! I love this website!

[55 x 54 x 53 x 52] / [4 x 3 x 2 x 1] =8,185,320 / 24 =341,055

Here's a pic for 1

AB is the pole = 21.13m, DC is the tower = 50m....BD = 50m..angle ECA = 30....angle ECB  = 45

Oops, tertre...

I meant to write

a = 7 , b = 4 and c  = 80 ....so

a + b + c   =  91

1. From the top of a tower of height 50m, the angles of depression of the top and bottom of a pole are 30° and 45° respectively. Find 

a) how far the pole is from the bottom of the tower

b) the height of the pole (use root3 =1.732)

a) The distance, D, is given by :

D = 50 / tan(45)  = 50m

b) The height of the pole, H, can be found by

H =  50 - 50tan(30)   =  50 - 50/sqrt(3)   =  50 - 50/1.732  ≈ 21.13 m

It says wrong, why?
 

OK! When did the EROICA premiere?

Very nice, MJ   ......an original ??

49x^2+56x-64 = 0

49x^2 + 56x  =   64  divide through by  49

x^2 + (56/49)x  =  64/49 

x^2 + (8/7)x  = 64/49

Take (1/2) of (8/7)   = 8/14  = 4/7   square it   = 16/49   add to both sides

x^2 + (8/7)x + 16/49 =  64/49 + 16/49   

x^2 + (8/7)x + 16/49 = 80/49    multiply through by 49

49x^2  + 56x + 16  = 80      factor the left side

(7x + 4)^2   = 80

a = 7, b = 4 , c = 8

a + b + c  =   19

*returning quest, posted first answer*

I enjoy this forum while I'm working. Some of the stuff posted gives me the giggles and brightens my day. I use the calc. at work and scroll through the questions just to see what's going on. Maybe learn something. Thanks, guys <3

What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594

\(\frac{180}{594}=\frac{2^2\cdot3^2\cdot5}{2\cdot3^3\cdot11}\ \ lcm=2^2\cdot3^3\cdot5\cdot11\ \ gcf=2\cdot3^2\)

\(\frac{lcm}{gcf}=\frac{2^2\cdot3^3\cdot5\cdot11}{2\cdot3^2}=\frac{2\cdot3\cdot5\cdot11}{1}\color{blue}=330:1\)

  !

Jup it is 56.5 km^2. At least referring to wikipedia. I actually dont need any correct answer, just a value that gets near the 550 Gw/h. I mean, the formula 0.5*A*p*h^2*g describes the maximal potential Energy in one high tide, meaning it cant get over this value. Either theres a mistake in my logic or calculation or the values given by wiki are wrong. only thing that would come to mind are the difference in high tides during the year between springtide and nipptide and difference in position between moon, sun and earth...

@Guest, Um, not necessary.

My apologies, is there anything I can do for you?

LCM[180, 594] =5,940.

GCD[180,594]  =18

Ratio =5,940 / 18 =330

Thanks so much!

x^2 + 2sqrt(3)x + 1  = 0

x^2 + 2sqrt(3)x  = - 1         

Take (1/2) of 2sqrt(3)  = sqrt(3)....square this = 3   add to both sides

x^2 + 2sqrt(3)x + 3 =  -1 + 3      simpify and factor

(x  + sqrt(3) )^2   =    2

r =  sqrt(3), s = 2

27sin^2x-3sinx=cos^2x  

27sin^2x - 3sin x - cos^2x = 0

27sin^2x - 3 sinx - (1 - sin^2x) = 0

28sin^2x - 3sin x - 1   = 0   factor as

(7sin x + 1) (4sin x - 1)  = 0

Set each factor to 0  and we have

7 sin x + 1   = 0     subtract 1 from both sides

7sin x  = -1    divide both sides by 7

sinx = (-1/7)     take the sine inverse

arctan (-1/7) = x  ≈ -8.2°  ≈ 351.8°  and ≈ 188.2°

4sin x - 1  = 0    add 1 to both sides

4sinx = 1    divide both sides by 4

sin x = 1/4 take the sine inverse

arcsin (1/4) = x  ≈ 14.5°  and ≈ 165.5° 

Solutions [ 14.5°, 165.5°, 188.2°, 351.8° ]