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Ne4ueva [31]
2 years ago
7

Rhombus ABCD, with side length 6, is rolled to form a cylinder of volume 6 by taping side AB to side DC. What is sin(angle ABC)?

Mathematics
1 answer:
FinnZ [79.3K]2 years ago
3 0

Answer:

Sin(ABC)=π/9

Step-by-step explanation:

Properties of a Rhombus:

1. All sides are equal

2. Opposite sides are parallel to each other

                                             

<em>Using SOHCAHTAO </em>

Attached shape

SinƟ = Opposite/Hypotenuse

Sin Ɵ = h/6, making h the subject of the formula

H=6SinƟ

Volume of a cylinder = πr2h where r is the radius and h is the height.

Finding the radius of the cylinder:

After joining point AB a circle is created with a circumference of 6 thus

C=2πr where C = circumference and r= radius. Inputting the values from the question, length of one side is the circumference

6=2πr, <em>making r the subject of the formula </em>

6/2π =2πr/2π

This leaves r=6/2π,<em> reducing to lowest terms </em>

r=3/π

 

Finding the value of h:

Volume of a cylinder = πr2h <em>where r is the radius and h is the height</em>.

Volume was given as 6 , r=3/π and h=6SinƟ

<em>Substituting into the equation for the volume of the cylinder </em>

V=πr2h

6=π(3/π)2(6SinƟ)  <em>removing the squared sign</em>

6=π(9/π2)(6SinƟ)   <em>removing brackets </em>

6= 54SinƟ/π making SinƟ <em>subject of the formula </em>

SinƟ=6π/54 <em>reducing to lowest terms </em>

SinƟ=π/9  therefore Sin(ABC)= π/9

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1. Let a; b; c; d; n belong to Z with n &gt; 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

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(k+m)n  =    (a+c)-(b+d)

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Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

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