Hi there!
We can use right-triangle trigonometry to solve.
We are given the HYPOTENUSE and ADJACENT sides, so we must use cosine in this instance.
cosθ = Adjacent/Hypotenuse
We can plug in what is given:
cos(28) = A/17
Solve for 'A':
17cos(28) = <u>15.01 ft</u>
1. The shape of cross-section is a circle.
2. The face parallel to ABCD is EFGH. Since this is a a rectangular shape,
A = L*H = 12*6 = 72 cm^2
3. The cross-section parallel to ABC is DEF with h = 12 ft, b= 5ft (where h is the height and b is the base of a right angled triangle).
Area, A = 1/2 *b*h = 1/2*5*12 =30 ft^2
4. Plane BDHF is a rectangle shape whose length is the diagonal of ABCD.
Diagonal BD = sqrt (AB^2+BD^2) = sqrt (8^2+7^2) = 10.63 cm.
Perimeter, P = 2(BD+DH) = 2(10.63+6) = 33.26 cm
Answer
The expression is equivalent to sin^{2}x
The first step is to multiply the two expressions between parentheses :
(II)
There is a trigonometric identity that states :
Working with this expression :
⇒
(I)
Using the equation (I) in (II) :
⇒
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Yes that’s correct
14 x 2 =28
28 - 5
=23
Answer by answer key by key question equals equation
