- (A) The mainsail foot is 13 feet long, the luff is 10 feet long, the leach is 6.93 feet long, and the mainsail area is 64.85 ft².
- (B) The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff forms a 50° angle with the leach at the head, requiring the length of the leach and the area of the mainsail.
<h3>
What exactly is sines law?</h3>
- The law of sines refers to the ratio of a triangle's side length to the sine of the opposite angle, which is the same for all sides.
- SinA/a = SinB/b = SinC/c
- Where A, B, and C represent the triangle's angle and a,b, and c represent the opposite sides of the triangle's angle.
So, Part A:
By employing the sines law,
- SinA/10 = SinB/13
- SinA/10 = Sin50/13
- A = 36.10
- C = 180-(180-36.10)
- C = 93.90
So,
- Sin36.10 / a = Sin50/ b = Sin93.90/c
- Sin36.10 / 10 = Sin50/ 13 = Sin93.90/c
- c = Sin93.90 × 13 ÷ Sin50
- c = 16.93 ft
Now, Part B:
Area of the mainsail:
- = 1/2 × abSinC
- = 1/2 × 10 × 13 × Sin93.90
- = 64.85
Therefore, the answers to both parts are:
- (A) The mainsail foot is 13 feet long, the luff is 10 feet long, the leach is 6.93 feet long, and the mainsail area is 64.85 ft².
- (B) The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff forms a 50° angle with the leach at the head, requiring the length of the leach and the area of the mainsail.
Know more about sines law here:
brainly.com/question/27174058
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The complete question is given below:
A sailmaker is designing a mainsail for a newly constructed sailboat. The mainsail foot is 13 feet long, the luff is 10 feet long, and the luff makes a 50° angle at the head with the leach.
Part A: Determine the length of the leach.
Part B: Find the area of the mainsail.
PLS HELP :) 50 POINTS!!!
Answer:
The probability will be 0.3085 or 0
Step-by-step explanation:
Given:
True mean=12.5
Sample mean =12.6
Standard deviation=0.2
Samples=100
To Find:
Probability that exceeds 12.6 ounces.
Solution:
Calculate the Z-score for given means and standard deviation.
So
Z-score= (true mean -sample mean)/standard deviation.
Z-score=(12.5 -12.6)/0.2
=-0.1/0.2
=-0.5
Now Using Z-table
P(X≥-0.5)=p(Z≥-0.5)=0.3085
Hence Probability that sample mean weight exceeds will be 0.3085
OR
By using Normal distribution with sampling ,it will be as follows
Z=(X-u)/[Standard deviation/Sqrt(No of samples)]
Z=(12.6-12.5)/(0.2/Sqrt(100)
Z=0.1/0.2/10
Z=5
So P(X≥12.6 )=P(Z≥5)=1
Pr(Z≥5)=1-1=0.
(Refer the attachment )
Hence Probability of getting ounces greater than 12.6 is '0'.
The sampling is of 0.02 size hence graphically it looks likely.
as shown in attachment.
Answer:
I think is A but I could be wrong.....But I don't think so because I did the math lol
Step-by-step explanation:
|AD|=(3i-1j)-(-3i-0j)=3i-1j+3i+0j=6i-1j
|BA|=(-3i-0j)-(-1i+3j)=-3i-0j+1i-3j=-2i-3j
Area =|AD|x|BA|
= i j k
6 -1 0
-2 -3 0
= -1 0 i - 6 0 j + 6 -1 k
-3 0 -2 0 -2 -3
= 0 i - 0 j + (-18-2)k = -20 k
√(-20)² =20
