Trigonometry word problems?

I am stuck on the following three problems any help is appreciated.

1. Two buildings with flat roofs are 60 feet apart. From the roof of the shorter building 40 ft in height the angle of elevation to the edge of the roof of the taller building is 40 degree. How high is the taller building.

2. A ladder with its foot in the street makes an angle of 30 degrees with the street when its top rests on a building on one side of the street and makes angle of 40 with the street when its top rest on a building on the other side of the street, if the ladder is 50 feet long how wide is the street.

• Step 1: Draw a picture; Step 2: Identify the triangle being described by the word problem; Step 3: Fill in all the information you are given. Step 4: Use trigonometric functions or identities to find the quantity being asked. – Arturo Magidin Jun 23 '12 at 19:18
• @AndréNicolas I think number 2 has enough information. I am just assuming the ladder is fixed someplace in the street. First it touches the first building and make some angle, then keeping the base, you flip onto the other side until it touches the other building. – William Jun 23 '12 at 19:20
• @William: Yes, it is a matter of interpretation. With not moving foot of ladder, and leaning against (quite tall) buildings, everything is OK. – André Nicolas Jun 23 '12 at 19:39

1. From the top of the shorter building to the top of the higher building, you can form a triangle with 40 degree angle at the tip of the shorter building with one leg 60 feet, the other leg is $x$ which represent how much higher is the other building. (Draw a picture yourself; or see Zev Chonoles picture above.) So you see that $\tan(40) = \frac{x}{60}$. So $x = 60\tan(40)$. So the height of the tall building is $40 + 60\tan(40)$.
2. For this problem, I am assuming where the ladder touches the street stays the same, i.e. you are you moving the top end to the other building. If you draw the picture, you see the distance on the street when the ladder touches the first first building is the length of the leg $x$, where $\cos(30) = \frac{x}{50}$, so $x = 50\cos(30)$. Similarly, when it touches the other building $\cos(40) = \frac{y}{50}$, where $y$ is the distance of the street. Hence $y = 50\cos(40)$. So the width of the sheet is the sum $50(\cos(30) + \cos(40))$.