Answer:
13. x = 52.8°
14. 663.65 ft
15. 30.7 ft
Step-by-step explanation:
The mnemonic SOH CAH TOA is intended to remind you of the relations between sides of a right triangle and trig functions of the acute angles.
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<h3>13.</h3>
The sides adjacent and opposite the angle are marked. The relevant trig relation is ...
Tan = Opposite/Adjacent
tan(x°) = 2.5/1.9
The angle is found using the inverse tangent function:
x° = arctan(2.5/1.9) ≈ 52.8°
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<h3>14.</h3>
Again, the tangent relation comes into play. The given values are the side opposite and the angle, and we are asked for the side adjacent.
Tan = Opposite/Adjacent
tan(11°) = (129 ft)/(distance to shore)
distance to shore = (129 ft)/tan(11°)
distance to shore ≈ 663.65 ft
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<h3>15.</h3>
In this scenario, the given angle is opposite the given side of the triangle. The measure of the hypotenuse is needed.
Sin = Opposite/Hypotenuse
sin(71°) = (29 ft)/(ladder length) . . . . substitute given information
ladder length = (29 ft)/sin(71°) . . . . . . solve for ladder length
ladder length ≈ 30.7 ft
(-2,3) to (2,-3)
6^2 + 4^2 = c^2
c = 7.21
————
(-2,3) to (1,5)
2^2 + 3^2 = c^2
c = 3.61
————
3.61 * 7.21 = approx. 26.03
The area is 26.03un^2
9.08+x=25.08
-9.08 -9.08
x=16
3/4x=12
x4/3 x4/3
x=16
4.5x=72
/4.5 /4.5
x=16
Both
Question:
Find the gradient of the line passing through (6,8) and (4,10).
Answer:
-1 is the right answer.
Step-by-step explanation:
Slope of the line = The gradient of the line
Gradient of the line is known as change in the value of y-axis by change in the value of x-axis
Gradient = ∆y\∆x
