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avatar+1442 

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

 

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

 

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

 

4:  How many values of x with \(0^\circ \le x < 990^\circ\) satisfy sin x = -0.31?

 

Thanks so much!

AnonymousConfusedGuy  Jun 21, 2018
 #1
avatar+91297 
+1

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

 

 

cool cool cool

CPhill  Jun 21, 2018
 #2
avatar+91297 
+1

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

 

 

cool cool cool

CPhill  Jun 21, 2018
 #3
avatar+91297 
+1

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 

 

 

cool cool cool

CPhill  Jun 21, 2018
edited by CPhill  Jun 21, 2018
 #4
avatar+91297 
+1

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°

 

 

cool cool cool

CPhill  Jun 21, 2018
 #5
avatar+1442 
+2

Thank you so much CPhill!

AnonymousConfusedGuy  Jun 21, 2018

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