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lesya692 [45]
3 years ago
6

In ΔIJK, k = 57 inches, i = 37 inches and ∠J=141°. Find ∠I, to the nearest degree.

Mathematics
1 answer:
Sonbull [250]3 years ago
8 0

Answer:

<I= 15degrees

Step-by-step explanation:

Using the cosine rule formulae;

j² = i²+k²-2i cos <J

j² = 37²+57² - 2(37)(57)cos <141

j² = 1369+ 3249- 4218cos <141

j² = 4618- 4218cos <141

j² = 4618-(-3,278)

j²= 7,896

j = √7,896

j = 88.86inches

Next is to get <I

i² = j²+k²-2jk cos <I

37² = 88.86²+57² - 2(88.86)(57)cos <I

1369 = 7,896.0996+ 3249- 10,130.04cos <I

1369 = 11,145.0996 - 10,130.04cos <I

1369 - 11,145.0996 = - 10,130.04cos <I

-9,776.0996=- 10,130.04cos <I

cos <I =9,776.0996 /10,130.04

cos<I = 0.96506

<I = 15.19

<I= 15degrees

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What is the translation from quadrilateral IJK to<br> quadrilateral I'J’K’
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