Please help. Thanks.

I can help with part 2. 

 

angle A is cos(16/20) and angle B is sin(12/20)

S  =  2r³t

                       To solve this equation for  t ,  first let's divide both sides by  2 .

\(\frac{S}{2}\)  =  r³t

                       Next divide both sides by  r³ .

\(\frac{S}{2r^3}\)  =  t

 

t  =  \(\frac{S}{2r^3}\)

 

---------------------------

 

4x² - 376  =  24

                                 To solve this equation for  x , first let's add  376  to both sides.

4x²  =  24 + 376

 

4x²  =  400

                                 Then divide both sides by  4 .

x²  =  400/4

 

x²  =  100

                                 Then take the ± square root of both sides.

x  =  ±√100

 

x  =  ± 10

 

This means there are two values of  x  that make the equation true,

and those two values are 10  and  -10 .

For the third question...

 

∠A  and  ∠B  must be acute angles because the third angle is 90°, and the sum of the angle measures in a triangle is 180°, and each angle has to have a measure greater than 0°. (If either  ∠A  or  ∠B  were  90°  then two of the sides would be parallel, and they could never meet.)

 

First let's find the sine and cosine of angle A .

 

sin A  =  opposite / hypotenuse

sin A  =  12 / 20

sin A  =  3 / 5

 

cos A  =  adjacent / hypotenuse

cos A  =  16 / 20

cos A  =  4 / 5

 

Next let's find the sine and cosine of angle B .

 

sin B  =  opposite / hypotenuse

sin B  =  16 / 20

sin B  =  4 / 5

 

cos B  =  adjacent / hypotenuse

cos B  =  12 / 20

cos B  =  3 / 5