Explanation:
Provided <u>2 length sides</u> and <u>one angle</u> also need to find <u>one missing side</u>.
So, use cosine rule:
a² = b² + c² - 2bc cos(A)
<h3><u>Part 1</u></h3>
c² = 9² + 11² - 2(11)(9) cos(57)
c² = 94.16147
c = √94.16147 = 9.70 cm
<h3><u>Part 2</u></h3>
d² = 5² + 7² - 2(5)(7) cos(48)
d² = 27.16
d = √27.16 = 5.21
<h3><u>Part 3</u></h3>
5² = 7² + 9² - 2(7)(9) cos(H)
-126cos(H) = 25 - 49 - 81
cos(H) = -105/-126
cos(H) = 5/6
H = cos⁻¹(5/6) = 33.56°
<h3><u>Part 4</u></h3>
8² = 4² + 7² - 2(4)(7) cos(J)
-56cos(J) = 64 - 16 - 49
cos(J) = -1/-56
J = cos⁻¹(1/56) = 88.98°
Passcode: 3142
Answer:
frick
Step-by-step explanation:
chicken jencdcdhchdbbchdcdccdc
Answer:
2n+2
Step-by-step explanation:
The pool can hold 65.84 ft³ of water
<u>Explanation:</u>
Given:
Shape of pool = octagonal
Base area of the pool = 22 ft²
Depth of the pool = 3 feet
Volume, V = ?
We know:
Area of octagon = 2 ( 1 + √2) a²
22 ft² = 2 ( 1 + √2 ) a²

a² = 
a² = 4.55
a = 2.132 ft
Side length of the octagon is 2.132 ft
We know:
Volume of octagon = 

Therefore, the pool can hold 65.84 ft³ of water
Since the angle
100
° is in the second quadrant, the reference angle formula is A
r
=
180
°
−
A
c. A
r
=
180
°
−
100
°
The reference angle is A
r
=
80
°
.
80
°