Answer:
AB = 75
BC = 60
AC = 45
m∠A = 53°
m∠B = 37°
m∠C = 90°
Step-by-step explanation:
<u>Trigonometric ratios</u>

where:
is the angle- O is the side opposite the angle
- A is the side adjacent the angle
- H is the hypotenuse (the side opposite the right angle)
Given:

Therefore:
- side opposite angle A = BC = 60
- side adjacent angle A = AC = 45
To find the length of AB (the hypotenuse), use Pythagoras’ Theorem:

(where a and b are the legs, and c is the hypotenuse, of a right triangle)
⇒ AC² + BC² = AB²
⇒ 45² + 60² = AB²
⇒ AB² = 5625
⇒ AB = √5625
⇒ AB = 75
To find m∠A:



m∠C = 90° (as it is a right angle)
The interior angles of a triangle sum to 180°
⇒ m∠A + m∠B + m∠C = 180°
⇒ 53° + m∠B + 90° = 180°
⇒ m∠B = 180° - 53° - 90°
⇒ m∠B = 37°
Answer: 2???????
It's too short. Write at least 20 characters to explain it well.
Here you go!! Hope it helps !!
Answer:
The area of the rectangle is increasing at a rate of 84 square centimeters per second.
Step-by-step explanation:
The area for a rectangle is given by the formula:

Where <em>w</em> is the width and <em>l</em> is the length.
We are given that the length of the rectangle is increasing at a rate of 6 cm/s and that the width is increasing at a rate of 5 cm/s. In other words, dl/dt = 6 and dw/dt = 5.
First, differentiate the equation with respect to <em>t</em>, where <em>w</em> and <em>l</em> are both functions of <em>t: </em>
![\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left[w\ell]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7BdA%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5Bw%5Cell%5D)
By the Product Rule:

Since we know that dl/dt = 6 and that dw/dt = 5:

We want to find the rate at which the area is increasing when the length is 12 cm and the width is 4 cm. Substitute:

The area of the rectangle is increasing at a rate of 84 square centimeters per second.
1/4 * 1/3 * 1/2 * 1/1 = 1/24 chance that all 4 letters are placed in the correct envelopes.
With the first envelope, there are 4 choices of letters, so the probability of picking the correct letter is 1/4.
With the second envelope, there are 3 remaining letters, so the probability of picking the correct letter is 1/3.
The same logic follows the the third and final letter.