+0  
 
0
242
4
avatar+124 

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

FiestyGeco  Jun 8, 2018
 #1
avatar+92344 
+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

CPhill  Jun 9, 2018
 #2
avatar+92344 
+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

CPhill  Jun 9, 2018
 #3
avatar+92344 
+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

CPhill  Jun 9, 2018
 #4
avatar+92344 
+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

CPhill  Jun 9, 2018

22 Online Users

avatar
avatar
avatar
avatar
avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.