Let the legs of the triangle be a and b, and the hypotenuse c.
Your first instinct might tell you to use the Pythagorean theorem to go about solving this because
![c= \sqrt{a^{2}+b^{2}}](https://tex.z-dn.net/?f=c%3D%20%5Csqrt%7Ba%5E%7B2%7D%2Bb%5E%7B2%7D%7D%20)
. This works, but it is slow.
The fastest way to solve this is to recognize that the right triangle is a special triangle where the ratio of the sides are 3:4:5. This means that if the legs are 9 and 12, then the hypotenuse is 15 because 3*3 is 9, 3*4 is 12, and 3*5 is 15.
Answer: The tailor made 8 complete hats
Step-by-step explanation:
Amount of fabrics available to make hat = 6 2/3 yards
1 hat requires = 5/6 yards
Number of hat to obtain from the available fabric = Amount of fabrics available to make hat / requirement of 1 hat
=(6 2/3) / (5/6)
(20/3) / (5/6)
20/3 x 6/5 = cancelling out, we are left with
4x2= 8 hats
or
(20/3) / (5/6)
6.6666 /0.83333
= 8.00002= 8 hats
The answer is b if you do the equation right
Answer:
You can use either of the following to find "a":
- Pythagorean theorem
- Law of Cosines
Step-by-step explanation:
It looks like you have an isosceles trapezoid with one base 12.6 ft and a height of 15 ft.
I find it reasonably convenient to find the length of x using the sine of the 70° angle:
x = (15 ft)/sin(70°)
x ≈ 15.96 ft
That is not what you asked, but this value is sufficiently different from what is marked on your diagram, that I thought it might be helpful.
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Consider the diagram below. The relation between DE and AE can be written as ...
DE/AE = tan(70°)
AE = DE/tan(70°) = DE·tan(20°)
AE = 15·tan(20°) ≈ 5.459554
Then the length EC is ...
EC = AC - AE
EC = 6.3 - DE·tan(20°) ≈ 0.840446
Now, we can find DC using the Pythagorean theorem:
DC² = DE² + EC²
DC = √(15² +0.840446²) ≈ 15.023527
a ≈ 15.02 ft
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You can also make use of the Law of Cosines and the lengths x=AD and AC to find "a". (Do not round intermediate values from calculations.)
DC² = AD² + AC² - 2·AD·AC·cos(A)
a² = x² +6.3² -2·6.3x·cos(70°) ≈ 225.70635
a = √225.70635 ≈ 15.0235 . . . feet