Expert: Sherman D. - 12/16/2006

Question

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The text above is a follow-up to ...

-----Question-----
Hey, can you help with me with these two questions i tried to solve but got them incorrect must show work.

1). Using exact values, find the value of:

12 sin 30° -6 tan 45°+ sec45°/52


2).In a right triangle sin A=3/4 and c=17. Find a and b be sure to draw a right triangle and letter it properly.


-----Answer-----
By this do you mean
12sin(30°) - 6tan(45°) + (sec(45°)/52)
or
12sin(30°) - 6tan(45°) + sec((45/52)°)

if you answer that, i will answer both.

12sin(30°) - 6tan(45°) + (sec(45°)/52)

thats it but i just typed it the way it looked on my paper

Answer
12sin(30°) - 6tan(45°) + (sec(45°)/52)

Assuming that you do mean that sec(45°) is over 52

Using a calculator

12(1/2) - 6tan(1) + ((1/cos(45))/52)

since sec(45) = 1/(cos(45)) = 1/(sqrt(2)/2) = 2/sqrt(2) = sqrt(2)

(sqrt(2)/52)

so

12sin(30°) - 6tan(45°) + (sec(45°)/52) = (sqrt(2)/52)

I hope you were allowed to use a calculator to work out both.

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sinA = (3/4)
A = 48°35'25.36"

Using that, you get

a = BC
b = AC

sinA = a/17
sin(48°35'25.36") = a/17
a = 17sin(48°35'25.36")
a = 12 3/4 or 12.75

cosA = b/17
cos(48°35'25.36") = b/17
b = 17cos(48°35'25.36")
b = about 11.24

If you were to use a^2 + b^2 = c^2 to check to see if the lengths work, you will find out that they do.