+0

0
222
3
+156

Aug 17, 2018

#1
+2

I can help with part 2.

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

Aug 17, 2018
#2
+7616
+3

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 .

Aug 17, 2018
#3
+7616
+3

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

Aug 17, 2018