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1)Let theta be an acute angle such that sin(theta) = 3/5. What is the value of cos(180-theta)?

 

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

 

3)In triangle GHI we have GH = HI = 25 and GI = 30. What is sin angle GHI?

 

4)Let A be an acute angle such that sin4A = SinA. What is the measure of A in degrees?

 

5)In the diagram below, we know tan theta = 3/4. Find the area of the triangle.

Picture: https://latex.artofproblemsolving.com/1/5/b/15b2f668fb8720085a0232cba109b53f90a980d0.png

 Jun 8, 2018
 #1
avatar+107475 
+1

1)Let theta be an acute angle such that sin(theta) = 3/5. What is the value of cos(180-theta)?

 

Note...cos theta  =  4/5

 

cos (180  - theta)  =

 

cos (180) cos (theta)   + sin (180) sin (theta)  =

 

-cos (theta)  +  (0) * sin (theta)  = 

 

-cos (theta) =

 

- 4/5

 

 

cool cool cool

 Jun 9, 2018
 #2
avatar+107475 
+1

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

 

                                          H

                                25               25

                             G        30             I

The triangle is isosceles

Draw altitude HA.......this will bisect GI

And, using the Pythagorean Theorem,, HA  = √[ HI^2 - (GI/2)^2 ] = √[ 25^2 - 15^2]  = √400  = 20

 

So.....the sine of angle GIH  =  HA / HI  =  20 / 25   =  4 / 5

 

 

 

cool cool cool

 Jun 9, 2018
 #3
avatar+107475 
+1

5)In the diagram below, we know tan theta = 3/4. Find the area of the triangle.

 

If the tan  = 3/4....then the sin  =  3/5

 

So....the area of the triangle is

 

(1/2) (40) (60) sin (theta)

 

(1/2)(40)(60)(3/5)  =

 

(3/10) (2400)  =

 

720 units^2

 

 

cool cool cool

 Jun 9, 2018
 #4
avatar+107475 
+1

4)Let A be an acute angle such that sin4A = SinA. What is the measure of A in degrees?

 

sin (4A)  =  sin A

sin ( 2 * 2A)  = sin A

2sin 2A cos 2A  = sin A

2 [ 2 sin A  cos A ] * [ 2cos^2 - 1 ]  = sin A

8sin A cos^3 A  - 4sin A cos A  - sin A  = 0

sin A [ 8 cos^3 A - 4 cos A - 1  ]  = 0

 

sin A  = 0   ⇒  A  = 0°    [reject....A  is  > 0° ]

 

8cos^3 - 4cos A  - 1   = 0 ....   let  cos A  = x

8x^3 - 4x  - 1   = 0

( 2x + 1) ( 4x^2 - 2x  - 1 )    = 0 

 

Setting the first factor to 0 and solving for x, we have that

x = -1/2  ⇒  cos A  = -1/2      ⇒ A = 120°  [reject....A  is acute ]

 

Setting the second factor to 0 and solving for x, we have

4x^2 - 2x - 1   =  0

The solutions to this are :

x = [ 1  - √5 ] / 4  ⇒  cos A  = [ 1 - √5] /4  ⇒  A  = 108°  [ reject....A  is acute ]

x = [ 1 + √5 ] / 4 ⇒  cos A  =  [ 1  + √5 ] / 4 ⇒ A  =  36°

 

 

cool cool cool

 Jun 9, 2018

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