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zzz [600]
3 years ago
13

Cos s=-2/5 and sin t=4/5, s and t are in quadrant II find cos(s+t) and cos(s-t)

Mathematics
1 answer:
mojhsa [17]3 years ago
5 0

Answer:

•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25

•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25

cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗

sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5

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See diagram
I wrote that the degrees are 70 each because it is isocolese and 140/2=70


a.
to solve for the diagonal, I'm going to use trig to find the height and the base then use pythagorean theorem to find the diagonal

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see I've drawn an auxiliry line to make a right triangle on the leftmost side
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sin(70°)=h/7
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b.
we got that the base was 7cos(70°)
2 of those bases so 22-14cos(70°)=17.2117ft or rounded 17.21ft




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b. 17.21ft

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