Airplane

Section 6.1













I messed up all parts except for (a). . .maybe.
Can you set up each part for me to try again?
I want to try a second time. After all, I need all the practice I can get.

 

you messed up (a) first, angle C=17 not 7

given:

<PAC = 55°

<PBC = 72°

AB = 2.2 mi

let the point where altitude touches extension of the AB be C

<ABP = 180 - <PBC = 180 - 72 = 108°

<APB = 180 - <BAP - <ABP = 180 -55 - 108 = 17°

 

Can you please draw the correct geometric figure as a guide for me to use? The angles you found pertain to which part of the problem? If not, set up each part for me to work on Sunday. Good night.

 

your drawing if good, just mark angle 17 degrees instead of 7

 

I will redraw the diagram.

 

Let me see.

Part (b)

a/sinA = c/sin C

asinC = csinA

asin17 = (2.2)(sin55)

a = (2.2)(sin55)/sin(17)

a = 6.2 miles

Part (c)

Dropping a perpendicular from angle C to D, I form a right triangle.

180° - 108° = 72°.

sin(72) = h/6.2

sin(72)(6.2) = h

5.89 miles = h

Part (d)

Let AC = b

b/sin(108) = 2.2/sin(17)

Solving for b, I get 7.2 miles from A to C.

I can now use the Pythagorean Theorem.

Let d = distance plane must travel before reaching A.

sin(35) = d/7.2

Solving for d I get 4.13 miles.

You say?

 

correct

 

I will post 3 SSA case law of sines. I then will post several applications. However, I don't want you to work out the word problems UNLESS indicated to do so. Just set it for me to work out. One problem now and the rest later tonight or tomorrow. My weekend is over just like that.