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September 10th, 2017, 11:35 AM   #1
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Trigonometry Homework Question

Hello, I'm given a right-angled triangle with an angle of 60 degrees, nothing else. The adjacent side is named x and the opposite side is named h. The question is 'write h in terms of x. I know that the answer will be h=(square root of 3)x, as I'm given the answers at the back of the book. However, I would like to know how to solve it. Any idea how?

Last edited by skipjack; September 10th, 2017 at 03:44 PM.
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September 10th, 2017, 02:51 PM   #2
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You need to use the cosine (1/2) of 60 deg.
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September 10th, 2017, 06:06 PM   #3
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Let A denote the vertex where the 60° angle is, B denote the vertex where the right angle is, and C denote the remaining vertex of the triangle.

Let D be the point such that ABCD is a rectangle and O be the point of intersection of AC and BD.

By symmetry, angle ABD = 60°, so triangle ABO is equilateral and AC = 2AO = 2x.

By Pythagoras, h² = AC² - AB² = 4x² - x² = 3x², so h = √3x.
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September 10th, 2017, 06:20 PM   #4
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tan(a) = h/x

solve for h,


h = tan(a)*x


But tan(a)=sin(a)/cos(a)

so h = (sin(a)/cos(a))*x

If you plug in 60 for a and fill in the steps you get h = sqrt(3)*x
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September 10th, 2017, 07:51 PM   #5
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How would you know the values of sin(60°) and cos(60°) (yet not tan(60°))?
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September 11th, 2017, 05:26 AM   #6
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Construct an equilateral triangle with sides equaling 1 unit. Construct an altitude that divides the triangle into to congruent right-angled triangles.
Then √(1 - (1/2)$^2$) = √3/2, so cos(60) = 1/2, sin(60) = √3/2 and tan(60) = √3.

tan(60) = h/x = √3, h = √3x.

Last edited by greg1313; September 11th, 2017 at 05:36 AM.
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