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A, B & C form the vertices of a triangle. CAB = 90°, ABC = 55° and AB = 9.4. Calculate the length of BC rounded to 3 SF.​

User Lcrepas
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1 Answer

1 vote
We can use the Pythagorean theorem to find the length of BC.
Let AC = x, then
x^2 = AB^2 + BC^2
x^2 = 9.4^2 + BC^2
x^2 = 88.36 + BC^2
Using the law of cosines,
x^2 = AB^2 + BC^2 - 2 * AB * BC * cos(ABC)
88.36 = 9.4^2 + BC^2 - 2 * 9.4 * BC * cos(55)
Solving for BC, we get:
BC = sqrt(88.36 + 9.4^2 - 2 * 9.4^2 * cos(55))
BC = sqrt(88.36 + 9.4^2 - 2 * 9.4^2 * (sqrt(2)/2))
BC = sqrt(88.36 + 9.4^2 - 2 * 9.4^2 * (sqrt(2)/2))
BC = sqrt(88.36 + 9.4^2 - 2 * 9.4 * 9.4 * (sqrt(2)/2))
BC = sqrt(88.36 + 9.4^2 - 2 * 9.4 * 9.4 * (sqrt(2)/2))
BC = sqrt(88.36 + 88.36 - 17.64 * (sqrt(2)))
BC = sqrt(176.72 - 17.64 * (sqrt(2)))
BC = sqrt(176.72 - 12.58)
BC = sqrt(164.14)
BC = 12.81
Rounding to 3 significant figures, BC = 12.81
User Jack Daniel
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