
Trig Word Problems
I have a test on Monday over simple trig functions. I can work out all of the equations but I am having a really tough time with the word problems. Any help would be great. i will post a few probs here and if anyone is willing to help beyond that please let me know. thanks in advance.
A ladder 8.0m long is placed against a building. How high will the top of the ladder reach the building? In the picture it gives me the ladder at 8.0m which is the hypotenuse, and it gives me the adjacent angle of the ladder to the ground at 61 degrees.
A helicopter hovers 800 ft directly above a small island that is off the california coast. from the helicopter, the pilot takes a sighting to a point O directly ashore on the mainland, at the water's edge. If the angle of depression is 35 degrees, how far off the coast is the island?

Hello,
First problem:
What you are looking for is the length of the vertical side.
Do you remember the definition in terms of sides of the sine ?

none of this was explained to me. you have to understand that I am only in Tech Math and they really don't go in depth on those things. the only thing they really give you is the formulas and tell you to plug it in. Of course out of the formulas I have been given I can not plug this in.

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I don't know what Tech class is :D
By the way, take a look at my drawing
sine, cosine and tangent are defined as following :
$\displaystyle \sin(\theta)=\frac ca=\frac{\text{opposit side}}{\text{hypotenuse}}$
$\displaystyle \cos(\theta)=\frac ba=\frac{\text{adjacent side}}{\text{hypotenuse}}$
$\displaystyle \tan(\theta)=\frac cb$

ok i had a brain fart and figured out question one! woohoo! can you please help with question 2?
Tech Math is a college course over Technical Mathematics. Basically what is does is takes all of the basic techniques in math and teaches it to one that really does not need to take the full courses for their degree.

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Ok, try solving by using my drawing :D

thanks so much. with the picture i already had and then the picture you gave me, i figured it out. the only different between the pic you gave and the pic i had was the pic i had showed the 35 angle on the outside of the triangle and you showed it inside. It is actually wrong to have it inside the triangle. the 35 degree angle was only part of a 90 degree angle. basically the pic i had showed 2 triangles. thanks for the help though. i will have more questions today. I have to do variations and system of equation word problems today.