Trigonometry (1 Viewer)

[12o9]

Just two trig questions I can't do for some reason . Thanks in advance.

uhh. Sorry about the quality of the scan, my school decided to supply me with a Fitzpatrick book which has been around since 1986. Oh yeah, the fact that i slightly moved the book whilst scanning doesn't help either. Thanks again.

Source - Fitzpatrick, Exercise 21 a)

 

[Dr. Zoidberg]

Labelling

Let AE = h = DF

Let EB = z

Let AD = x

EF = x - z (use subtraction)

angle EBA = 90 - θ (angle sum of straight line)

Therefore, angle BAE = θ (angle sum of triangle)

Angle EAD is a right angle by the way we constructed it, therefore, angle BAD = 90-theeta

Therefore, angle BDA = θ (angle sum of triangle)

Calculations

Area of rectangle AEFD = xh

Area of triangle AEB = 1/2 x h x z
= (hz)/2

Area of triange BDF = 1/2 x (x-z) x h
= (xh - zh)/2

Area of triangle ABD = area of rectangle AEFD - Area of triangle AEB - Area of triange BDF
= xh - (hz)/2 - (xh - zh)/2
= (2xh - hz - xh +zh )/2
= (xh)/2

now, by pythagoras' theorem, BD = sqrroot(x^2 - a^2)

Using trig,

cot θ = [sqrroot(x^2 - a^2)]/a

acot θ = sqrroot(x^2 - a^2)

Square both sides now

a^2 cot^2 θ = x^2 - a^2

Bring a^2 to the left

[a^2 cot^2 θ] + a^2 = x^2

a^2[cot^2 θ + 1] = x^2

a^2[cosec^2 θ] = x^2

Square root everything

therefore, x = acosec θ

Then we find the value of h which is:

cos θ = h/a

therefore, h= acos θ

We then sub h and x into Area of triangle ABD

Therefore, the Area of triangle ABD = {[acosec θ] x [acos θ]}/2
= {[a/sin θ] x acos θ}/2
= [a^2cot θ]/2

When the tank is back to horizontal, the area of the water is [a^2cot θ]/2
which is in the form of a rectangle.

Since the area of this rectangle is breadth x width, we divide this area by b
which gives us the final depth of [a^2cot θ]/2b

Hopefully my writing is understandable

 

[12o9]

Labelling

Let AE = h = DF

Let EB = z

Let AD = x

EF = x - z (use subtraction)

angle EBA = 90 - θ (angle sum of straight line)

Therefore, angle BAE = θ (angle sum of triangle)

Angle EAD is a right angle by the way we constructed it, therefore, angle BAD = 90-theeta

Therefore, angle BDA = θ (angle sum of triangle)

Calculations

Area of rectangle AEFD = xh

Area of triangle AEB = 1/2 x h x z
= (hz)/2

Area of triange BDF = 1/2 x (x-z) x h
= (xh - zh)/2

Area of triangle ABD = area of rectangle AEFD - Area of triangle AEB - Area of triange BDF
= xh - (hz)/2 - (xh - zh)/2
= (2xh - hz - xh +zh )/2
= (xh)/2

now, by pythagoras' theorem, BD = sqrroot(x^2 - a^2)

Using trig,

cot θ = [sqrroot(x^2 - a^2)]/a

acot θ = sqrroot(x^2 - a^2)

Square both sides now

a^2 cot^2 θ = x^2 - a^2

Bring a^2 to the left

[a^2 cot^2 θ] + a^2 = x^2

a^2[cot^2 θ + 1] = x^2

a^2[cosec^2 θ] = x^2

Square root everything

therefore, x = acosec θ

Then we find the value of h which is:

cos θ = h/a

therefore, h= acos θ

We then sub h and x into Area of triangle ABD

Therefore, the Area of triangle ABD = {[acosec θ] x [acos θ]}/2
= {[a/sin θ] x acos θ}/2
= [a^2cot θ]/2

When the tank is back to horizontal, the area of the water is [a^2cot θ]/2
which is in the form of a rectangle.

Since the area of this rectangle is breadth x width, we divide this area by b
which gives us the final depth of [a^2cot θ]/2b

Hopefully my writing is understandable

Massive thanks . Must have taken you ages to type that all out ._. , thanks again really appreciate it.

 

[u-borat]

man i hate to say it zoidberg, but theres a far simpler way man.

using zoidberg's diagram, there is a point D.

so now we want BD in terms of a and @.

now in that triangle BDA, tan @= a/BD
therefore BD=acot@
therefore area of triangle BDA= 1/2(acot@ X a)
now divide that by b and you have the height of the water.

 

[azureus88]

lol nice

 

[Dr. Zoidberg]

man i hate to say it zoidberg, but theres a far simpler way man.

using zoidberg's diagram, there is a point D.

so now we want BD in terms of a and @.

now in that triangle BDA, tan @= a/BD
therefore BD=acot@
therefore area of triangle BDA= 1/2(acot@ X a)
now divide that by b and you have the height of the water.

lol, damm, didnt realise

 

[hopethisworks]

Lol.

Lol.

fucking lol.

 

[lyounamu]

man i hate to say it zoidberg, but theres a far simpler way man.

using zoidberg's diagram, there is a point D.

so now we want BD in terms of a and @.

now in that triangle BDA, tan @= a/BD
therefore BD=acot@
therefore area of triangle BDA= 1/2(acot@ X a)
now divide that by b and you have the height of the water.

hahaha...lol

you ruined the party.