Question

need help with math problems

Answer 1

thank you for helping me i will give u lots of metals :)

Answer 2

Is it both questions or just the second one?

Answer 3

2nd one

Answer 4

OK I will oput some explanations down to help you understand.

Answer 5

nancy is helping me thanks gianfranco

Answer 6

i think she is

Answer 7

nancy i need help with both..i got confused i thought this was the other paper

Answer 8

Get back to me if you need to then )

Answer 9

thank u

Answer 10

frist find cost B
CosB =3/7
B = 64.6 degree
angle A=180-(90+64.6)=25.4 degree
tan B=X/3
X=3tan64.6
X=6,317
tan \[\tan A=\frac{3}{6,137}\]
A =26 degree
I don't know u need solve for angle or not

Answer 11

no that is good enough as long as i know the basic

Answer 12

R u sure?

Answer 13

let me done onr more

Answer 14

ok thank u

Answer 15

24)
BC=1/2 BA
BC=12:2= 6
\[AC=6\sqrt{3}\]\[Sin. B =\frac{6\sqrt{3}}{12}=\]\[Tan. B =\frac{6\sqrt{3}}{6}\]
I think u can use caculator so.ve very easy

Answer 16

thank u nancy u been a great help

Answer 17

good cluck your home work bye

Answer 18

bye

Answer 19

From the photo7.jpg:\[\text{AC}=10 \text{Cot}[50\text{Degree}]= 8.391 \]\[\text{AB}=10 \text{Csc}[50\text{Degree}]= 13.0541 \]To check the solutions:\[\sqrt{\text{AC}^2+10^2}= 13.0541 \]Note: Mathematica was used for the calculations. 50Degree is converted to the equivalent radian angle measure or 50/180 * pi. For example:\[10 \text{Csc}\left[\pi \left(\frac{50}{180}\right)\right]= 13.0541 \]