CPhill

In right triangle DEF, we have angle E = 90 degrees and tan D = 3*sin D. What is sin F?

tan D  = 3 sin D

sin D  / cos D  =   3 sin D        

Since  0° < D < 90°, then sin D  >  0

So...divide both sides by sin D and we have that

1 / cos D  = 3  .......rearrange as

cos D   = 1/3

Since D and F  are  complementary, then

cos D  = 1/3  =  sin  F

4.

In one period, the sine will be negative  from   180°  to 270°   and from 270° to 360°

So.....  360 + 360  + 270  = 990°

So....the sin x  = -0.31  =  [ 2 + 2  + 1 ]  =   5 times  from 0°  to 990°

We need to find the radius of the Earth  as follows :

25100  = 2 pi * radius

25100 / [ 2 pi]  = radius

12550 / pi   =  radius  [in miles]

At  45°,  the radius of the smaller circle parallel to the equator running through Lena will be =

12550  /  [ √2 pi]  miles

So....the circumference  of this smaller circle will be   =

2 pi   [ 12550 / (√2 pi ) ] =

√2 * 12550  ≈  17,748 miles  = 17,700 miles

3:  ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of line CD, then what is cos (angle AMB)?

Let AB  =  side of the tetrahedron  = s

Then BM =  AM  =   s * sqrt(3)/2

Using the Law of Cosines, we have that

AB^2  = BM^2 + AM^2  -  2 (BM)(AM) cos AMB

s^2  =   (3/4)s^2 + (3/4)s^2  - 2 ( 3/4)s^2 cos AMB

s^2  = (3/2)s^2  - (3/2)s^2  cos AMB

- (1/2)s^2   =  -(3/2)s^2  cos AMB

[ 1/2]s^2  / [ (3/2) s^2 ]  = cos AMB

[ 1/2]  / [3/2] = cos AMB

(1/2)( 2/3) = cos AMB

1/3   = cos AMB 

2:  In triangle GHI, we have GH = HI = 25 and GI = 30. What is sin (angle GHI)?

                                       H

                              25             25

                             G       30          I

This ttrangle is isosceles

Using the Law od Cosines

30^2   =  2 (25)^2 - 2(25)^2  cos A

-350 / [ -1250 ]  = cos A

cos A  =  7/25

sin A  = sqrt  [ 1 - cos^2 A ]  = sqrt [ 1  - (7/25)^2 ]  = sqrt [ 625 - 49] / 25  = 

sqrt [ 576] / 25   =    24 /25

1:  If A is an acute angle such that tan A + sec A = 2, then find cos A.

tan A  + sec A   = 2

sin A / cos A  +   1 / cos A  = 2

(sin A + 1 ) /cos A   =  2   ......  since A  is acute, then A < 90...so cos A is not 0

So ...multiply both sides by cos A

sin A + 1  = 2 cos A          square both sides

sin^2A + 2sinA + 1 = 4cos^2 A

sin^2A + 2sinA + 1  = 4(1- sin^2 A) ..... simplify

5sin^2 A + 2sin A - 3  =  0     factor

(5 sin A  - 3 ) ( sin A + 1)  =  0

So....we have two possible solutions

5  sinA - 3 =  0               sin A + 1  = 0

5 sin A  = 3                     sin A  = -1

sin A  = 3/5                     A  =  270°    (reject)

Since the sin A  = 3/5....then cos A  =   sqrt [  1 - ( 3/5)^2 ]  = sqrt [ 25 - 9] / 5  =

sqrt (16)/5   =   4/5

By the Triangle Inequality, the sum of any two sides is greater than the remaining side

So we have these inequalities :

11 + n  > `17      subtract  11 from both sides

n  >  6

And

11 + 17  > n    

28 > n  ⇒    n < 28

So....combining these, we have

6 < n < 28

So......the number of possible integer lengths for the remaining side  =

[28  - 6]  - 1  = 

22 - 1  =  

21

Do you want the graph of this  ???

Assuming that we cannot repeat any natural number....

10       ( 1 way)

9, 1     ( 2 ways)

8, 2     (2 ways)

7, 3     (2 ways)

6, 4     (2 ways)

7, 2 , 1   (6 ways)

6, 3 , 1   (6 ways)

5, 4, 1   (6 ways)

4, 3, 2 , 1    (24 ways)

So....   1 + 4(2)  + 3(6)  + 24    =     51 ways

I think the answer to the first question is  20 ways.....

To see this.....let Anne occupy one position in a circle.....and we will fill the sets in a  clockwise manner,.....  (we could also choose a counter-clockwise manner.....the results will be the same )

To Anne's right, we can choose any of the other 5 classmatres to fill this position.....and to this person's right, we can choose any 4 of the remaining classmates.....and these three comprise one set.....so, by default.....the other set will be composed of the classmates we haven't chosen

So......5 ways  X 4 ways  = 20 ways to divide the class