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Sholpan [36]
3 years ago
8

What is the approximate angle measure for angle W in the triangle below?

Mathematics
1 answer:
Artyom0805 [142]3 years ago
3 0
In this case we know the three sides of the triangle, then this is a SSS triangle (Side Side Side). To solve this case, first we must use the Law of Cosines, applied to the opposite side to the angle we want to find.
We want to find angle W, and its opposite side is XV, then we apply the Law of Cosines to the side XV:
XV^2=XW^2+WV^2-2(XW)(WV)cos W
Replacing the known values:
116^2=96^2+89^2-2(96)(89)cos W
Solving for W
13,456=9,216+7,921-17,088 cos W
13,456=17,137-17,088 cos W
13,456-17,137=17,137-17,088 cos W-17,137
-3,681=-17,088 cos W
(-3,681)/(-17,088)=(-17,088 cos W)/(-17,088)
0.215414326=cos W
cos W = 0.215414326

Solving for W:
W= cos^(-1) 0.215414326
Using the calculator:
W=77.56016397°
Rounded to one decimal place:
W=77.6°

Answer: Third option 77.6°
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Answer:

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B(0,2)\to B''(1,0)

C(1,2)\to C''(1,1)

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Step-by-step explanation:

The given trapezoid has vertices at A(−1,4), B(0,2), C(1,2) and D(2,4).

The transformation rule for 90° counterclockwise rotation is

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This implies that:

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B(0,2)\to B'(-2,0)

C(1,2)\to C'(-2,1)

D(2,4)\to D'(-4,2)

This is followed by a translation 3 units to the right.

This also has the rule: (x,y)\to (x+3,y)

A'(-4,-1)\to A''(-1,-1)

B'(-2,0)\to B''(1,0)

C'(-2,1)\to C''(1,1)

D'(-4,2)\to D''(-1,2)

Therefore:

A(-1,4)\to A''(-1,-1)

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D(-2,4)\to D''(-1,2)

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