Mathematics: Post your doubts here!

[RoOkaYya G]

Questions realted to Trignometry A2 ..
Q40 the underlined part
Q44 (ii) b
Q45
Q46

Answers :

ZaqZainab
Thought blocker
RoOkaYya G

40)
R = square root ( 1^2 + 1 ^2) = square root 2

alpha :
tan alpha = 1/1
alpha=tan inverse 1
= pi/4

Rcos (theta + alpha) = square root 2 (theta + pi/4) <----------ANSWER

cos theta + 1/ root 2 = sin theta
cos theta- sin theta = - 1 / root 2
Rcos (theta + alpha) = - 1 / root 2
square root 2 (theta + pi/4) = - 1 / root 2
solve for theta :
cos (theta+ pi/4) = - 0.5
(theta+ pi/4) = cos inverse 0.5
(theta+ pi/4) = pi/3
cos is negative so in quadrant S and T
in quadrant S (theta+ pi/4) = (pi -pi/3) = 2pi/3
in quadrant T (theta+ pi/4) = pi + pi/3 = 5 pi/3
theta = (2pi/3 - pi/4) and (5pi/3- pi/4)
=5pi/ 12 and 14pi/12
for acute angle, theta = 5pi/12 <---------------- ANSWER

44) ii) b)

solving part i) ull get it in R cos form as :
25sin(theta +73.7)....using this derive 25 cos (theta -73.7)

25 cos (theta-73.7) = -3
cos (theta-73.7) = -3/25
(theta-73.7) = cos inverse (3/25)
since cos is negative the solutions r found in S andT quadrant
in S, (theta-73.7) = 180 -83.1 = 96.9
in T,(theta-73.7) =180+ 83.1 = 263.1
(theta-73.7) =96.9 and 263.1

theta =(96.9+73.7) and (263.1 +73.7)
theta= 170.6 and 336.8
theta = 170.6 for 0< theta< 180 <-----------ANSWER

45)
equate the give equation of tan theta to 1/5
cross multiply

ull get x^2 +384 =5(8x)
x^2 + 384-40x =0
x^2 -40x +384 = 0

solving the quadratic :

(x-16)(x-24) =0
x= 16 and x=24 <------------ANSWER


ill post 46 in a while dinner

 

[nadeem siddique]

How p42?? Mechanics

 

[RoOkaYya G]

Haya Ahmed
taking the horizontal section of the corridor :
W = xcos theta + b sin theta
adjacent =cos theta * hypotenus

=xcos theta
opposite of smaller triangle = sin theta * b

=bsin theta

w= xcos theta + bsin theta (HENCE SHOWN)

the side W = the adjacent side of the line AD (when taken in a triangle) + the opposite side of angle theta
so :
calculating the the adjacent side of the line AD (when taken in a triangle) :
the adjacent side of the line AD (when taken in a triangle)
cos theta =adjacent/ hypotenus
adjacent=cos theta * hypotenus =cos theta * b
=bcos theta

calculating the opposite side of angle theta :
sin theta = opposite/ hypotenus
hypotenus = length a of rectangle - part x of side CD
= a-x
sin theta = opposite / (a-x)
opposite = (a-x) *sin theta


HENCE , w= b cos theta + (a-x) sin theta <--------- ANSWER

ii)
im unable to do ths part :S sorry

iii)
replace 45 - phi in the equation in part (ii) which is already given
use compound angle to expand and reach the solution asked there

iv) replace the values given in the equation given in part (iii)
solve for theta

ina hurry so plz manage
if u still dont get it then tell me for part iii and iv
ill do it stepwise

for (ii) im missing smethng i thnk. ill try it again later if u still dont get it. try ask someone else

 

[MazenAlosali]

Anyone have answers for paper 42 mechanics ??

 

[bludlynsolja]

no please!! lol i dont!
how was it btw?

 

[asadalam]

how to do q 20 guys?

 

[RoOkaYya G]

View attachment 48121 how to do q 20 guys?

is the answer 0.049??

 

[asadalam]

is the answer 0.049??

Yup,its 3/20pi,which is more or less the same as 0.049.

How?

 

[RoOkaYya G]

Yup,its 3/20pi,which is more or less the same as 0.049.

How?

dV/dx =dV/dt * dt/dx

dV/dx = (18*pi*x^2) - (4/3 * pi* x^3)

dV/dt = 3

we've to find dx/dt

replce the value of x in dV/dx by 2

using the chain rule, replace value obtain for dV/dx and the dV/dt.
ull get dt/dx = 194/9
dx/dt = 9/194 = 0.049 <------------ANSWER

 

[RoOkaYya G]

Yup,its 3/20pi,which is more or less the same as 0.049.

How?

its wrong my method -_________- i asked if its 0.049 n 3pi/20 makes 0.47
i missed a detail i guess...
:S

 

[asadalam]

its wrong my method -_________- i asked if its 0.049 n 3pi/20 makes 0.47
i missed a detail i guess...
:S

No 3/20pi does make 0.0477

ITs not 3pi its 3/20PI
How did u get the dV/dx.and where to use that radius of 6?

 

[RoOkaYya G]

No 3/20pi does make 0.0477

ITs not 3pi its 3/20PI
How did u get the dV/dx.and where to use that radius of 6?

oh ok..yeah u r right.
actually thts wht i thought i missed coz i dint use 6
the radius indicates the greatest depth of the container actually..

 

[RoOkaYya G]

No 3/20pi does make 0.0477

ITs not 3pi its 3/20PI
How did u get the dV/dx.and where to use that radius of 6?

differenciate the equation given...its dV/dx

 

[RoOkaYya G]

wait a min something is wrong

i ate the pi!!
the method is wrong somewhere. wait ill re do it

 

[asadalam]

oh ok..yeah u r right.
actually thts wht i thought i missed coz i dint use 6
the radius indicates the greatest depth of the container actually..

We dont have to use 6 anywhere in the working do we?
But by differentiation we get

1st step,simplifying:18pix^2 - pix^3/3
Differentiation of
dV/dx= 36pix - 3pix^2
How do you get that one?Dont we eliminate the constant below that is 3?And from where did you get that 4/3?

 

[RoOkaYya G]

We dont have to use 6 anywhere in the working do we?
But by differentiation we get

1st step,simplifying:18pix^2 - 4pix^3/3 <------we get 4 up there
Differentiation of
dV/dx= 36pix - 4pix^2 <---------- u get this when u re differenciate the one above
How do you get that one?Dont we eliminate the constant below that is 3?And from where did you get that 4/3?

 

[RoOkaYya G]

u r messing up my mind!! !

u said simplification??
sinplification gets u 6 pi x^2 - 4/3 pi x^2

 

[asadalam]

But where did the 4 come from????? isnt it just pix^3?

 

[asadalam]

u r messing up my mind!! !

u said simplification??
sinplification gets u 6 pi x^2 - 4/3 pi x^2

It was the whole devided by 3 ,further simplification would get you 6pix^2 as it would give 18/3.I forgot that.But where is the 4 coming from???

 

[RoOkaYya G]

But where did the 4 come from????? isnt it just pix^3?

i saw jst ur post n corrected it there.
wait lets take it from beginning! ok?

OMG!!! im so sorry!! i misread. its x^2 in the equation i used x^3

nxt time plz put ur attachments upright -_______- thank u