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Trig question help? Paper 1

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1) First identify equation(s) you should use. Area of triangle points to A = 1/2 a * b * sin C

From here we get 5 = 1/2 * 2 sqrt(5) * x * sin 2θ.
2) There are two unknowns, so we need one more equation. We are given only sin θ, so let's try to apply the double angle identity for sine.

sin 2θ = 2 sinθ * cosθ.

3) We seem to be getting closer, now we just need cosθ to find x. Use another identity sin²θ + cos²θ = 1. We see that θ is acute so cos θ is positive,

(2/3)² + cos²θ = 1 ===> cos θ = sqrt(5)/3

4) We have everything to solve for x.

5 = 1/2 * 2 sqrt(5) * x * 2 (2/3) sqrt(5)/3 Rearrange to get

x = 9/4

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On 02/05/2017 at 9:50 PM, kw0573 said:

1) First identify equation(s) you should use. Area of triangle points to A = 1/2 a * b * sin C

From here we get 5 = 1/2 * 2 sqrt(5) * x * sin 2θ.
2) There are two unknowns, so we need one more equation. We are given only sin θ, so let's try to apply the double angle identity for sine.

sin 2θ = 2 sinθ * cosθ.

3) We seem to be getting closer, now we just need cosθ to find x. Use another identity sin²θ + cos²θ = 1. We see that θ is acute so cos θ is positive,

(2/3)² + cos²θ = 1 ===> cos θ = sqrt(5)/3

4) We have everything to solve for x.

5 = 1/2 * 2 sqrt(5) * x * 2 (2/3) sqrt(5)/3 Rearrange to get

x = 9/4

Would this be a non-calculator question?

If not, we could do this without using the double angle identities, right?

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2 hours ago, Befuddled said:

Would this be a non-calculator question?

If not, we could do this without using the double angle identities, right?

Oh it would take me a while to find out. This look like paper 1 trig because that seems to be what is being tested out of this question. Double angle involvng a calculation often appears on paper 1 because without a calculator things simplify greatly.

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