Question

1. You have a 20 foot long ladder that you want to lean against a vertical wall. You want the top of the ladder to touch the wall 19 feet off the ground. What angle will the ladder form with the ground?
2. Using the same 20 foot ladder against the same vertical wall, you decide that it would be better if the ladder formed a 70 degree angle with the ground. How far up the wall will the top of the ladder reach?

3. You want to find the area of triangle BCD but all you have is the information provided in the image below. Then you realize you can use the special right triangle (30-60-90) to find the height of the triangle. Once you know the height and provided base measurement calculate the area.
What is the area of the triangle? Show all steps.

Answer 1

Answer:

1. The ladder forms 71.8° with the ground.

2. The top of the ladder will reach 18.79 feet up the wall.

3. Height = 8.66 cm and area = 21.65 sq. cm.

Step-by-step explanation:

1. If the angle of elevation of the ladder is $\theta$ then we can write  

$\sin \theta = \frac{\textrm {Perpendicular}}{\textrm {Hypotenuse}} = \frac{19}{20}$

⇒ $\theta = \sin ^{-1}(\frac{19}{20}) = 71.8$ Degrees.

Therefore, the ladder forms 71.8° with the ground. (Answer)

2.  Now, if the ladder formed a 70 degree angle with the ground and the length of the ladder remains the same as 20 feet, then we can write  

$\sin 70 = \frac{\textrm {Perpendicular}}{\textrm {Hypotenuse}} = \frac{x}{20}$

⇒ x = 20 sin 70 = 18.79 feet.

Therefore, the top of the ladder will reach 18.79 feet up the wall.

3. See the attached figure.

We have, $\tan 60 = \frac{BC}{CD} = \frac{BC}{5}$

⇒ Height = BC = 5 tan 60 = 8.66 cm.

Therefore, the area of the triangle BCD will be = $\frac{1}{2} \times CD \times BC = \frac{1}{2} \times 5 \times 8.66 = 21.65$ sq. cm. (Answer)