Find the length X | Geometry Problem | Important Geometry and Algebra Skills Explained

Find the length X | Geometry Problem | Important Geometry and Algebra Skills Explained

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[spacer]

Another method is to start with 2 equations:
Eq1: x = b cos(2θ) = a cos(θ) = a cos(2θ-θ) =
Eq2: x = b cos(2θ) = c cos(3θ) = c cos(2θ+θ) =
Since cos(-θ)=cos(θ) and sin(-θ)=-sin(θ), you can multiple Eq1 by c and Eq2 by a and then sum them to cancel the sin(2θ)sin(θ) term to give:
(ab+bc) cos(2θ) = 2ac cos(2θ)cos(θ)
The term cos(2θ) cancels, leaving x = a cos(θ) = (ab+bc)/2c

[jimlocke]

If the correct multiple choice answer is true for all values of Θ, it must be true for Θ=15°. In that case, ΔBCE is a well known 15°-75°-90° triangle, and the ratio of hypotenuse to long side is √6-√2, so a = (√6-√2)x. ΔBCD is a 30°-60°-90° special triangle and the ratio of hypotenuse to long side is (2√3)/3, so b = (2√3)x/3. ΔBCA is a 45°-45°-90° special triangle and the ratio of hypotenuse to either side is √2, so c = x√2. We try each of the multiple choice answers (A) x = ab+ac, (B) x = (ab+ac)/2c, (C) x = ab+bc, (D) x = (ab+bc)/2c. We can immediately rule out (A) and (C), because the right side will be a constant times x² and we need the right side to reduce to x. As a first cut, to avoid having to do all the arithmetic with radicals, use a scientific calculator to compute a, b, and c: a = 1.035276180x, b = 1.154700538x and c = 1.414213562 and then (B) and (D). So, in (B), x = 1.880575642x and in (D), x = So, we can rule out (B) and we need to evaluate the radicals in (D) to prove an exact equality. In (D), x = b(a+c)/2c = = ((2√3)x/3)((√6)x)/(2x√2) = ((2√3)x/3)(√6)/(2√2) = ((2√18)x/3)/(2√2) = ((2(3√2))x/3)/(2√2) = ((2√2)x)/(2√2) = x. So, x = x and (D) is correct.

This is a multiple choice question, so, unless the exam has an error, one of the answers must be correct, and (A), (B) and (C) are ruled out. If there were a fifth choice, "None of the above", we haven't proved that (D) is always true, so we would be forced to provide Math Booster's solution or something equivalent!

[pwmiles]

Tough one! You can work back from the choices offered i.e.

Put u = cos(θ). By standard identities you get
cos(2θ) = 2u^2 - 1
cos(3θ) = 4u^3 - 3u

We have
a = x/cos(θ)
b = x/cos(2θ)
c = x/cos(3θ)

Putting these into (D)
b(a+c)/(2c) = (x/(2u^2)-1))(x/u + x/(4u^3 -3u))/(2x/(4u^3-3u))
= (x/(2(2u^2 - 1))(1/u + 1/(4u^3 -3u))(4u^3-3u)
= (x/(2(2u^2 - 1))(4u^2 - 3 + 1)
= (x/(2(2u^2 - 1))(4u^2 - 2)
= x

[labzioui]

Since it is only asked to find the value of "x", the 2 data "a and b" are sufficient for this.
Cosθ = x/a
Cos2θ= x/b =2(Cosθ)^2-1

[robertbourke]

Very clever. I didn't get this one