I have a simple right angle triangle. All I am given is h (the hypotenuse) and that ratio of x:y is 2:3. What is the formula to find x and y in terms of h?
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Assignment Expert
16.10.14, 19:20
Dear James. Thank you for adding information. As to the first remark,
rewrite
x[sup]2[/sup]+(3x/2)[sup]2[/sup]=x[sup]2[/sup]+9x[sup]2[/sup]/4=x[sup]2[/sup](1+9/4)=(4/4+9/4)x[sup]2[/sup]=x[sup]2[/sup]*13/4.
We think our method is simpler. As to the second remark, we would say
that all arguments will work. The only thing left is to provide
several formulas: sin(atan(z))=z/sqrt(1+z[sup]2[/sup]),
cos(atan(z))=1/sqrt(1+z[sup]2[/sup]), where z=2/3. The answer will be
the same.
James
15.10.14, 11:58
Thank you for your answer, I will test it out. I am not sure how you
went from x^2 + (3/2 x)^2 to 13/4 x^2 I did come up with another
formula which I think works also: tan(θ) = opp/adj tan(θ) = x/y ∴
tan(θ) = 2/3 ∴ θ = atan(2/3) sin(θ) = opp/hyp sin(θ) = x/h ∴ x
= sin(θ).h {substitute θ with atan(2/3)} ∴ x = sin(atan(2/3)).h
cos(θ) = y/h ∴ y = cos(θ).h {substitute θ with atan(2/3)} ∴ y =
cos(atan(2/3)).h
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Dear James. Thank you for adding information. As to the first remark, rewrite x[sup]2[/sup]+(3x/2)[sup]2[/sup]=x[sup]2[/sup]+9x[sup]2[/sup]/4=x[sup]2[/sup](1+9/4)=(4/4+9/4)x[sup]2[/sup]=x[sup]2[/sup]*13/4. We think our method is simpler. As to the second remark, we would say that all arguments will work. The only thing left is to provide several formulas: sin(atan(z))=z/sqrt(1+z[sup]2[/sup]), cos(atan(z))=1/sqrt(1+z[sup]2[/sup]), where z=2/3. The answer will be the same.
Thank you for your answer, I will test it out. I am not sure how you went from x^2 + (3/2 x)^2 to 13/4 x^2 I did come up with another formula which I think works also: tan(θ) = opp/adj tan(θ) = x/y ∴ tan(θ) = 2/3 ∴ θ = atan(2/3) sin(θ) = opp/hyp sin(θ) = x/h ∴ x = sin(θ).h {substitute θ with atan(2/3)} ∴ x = sin(atan(2/3)).h cos(θ) = y/h ∴ y = cos(θ).h {substitute θ with atan(2/3)} ∴ y = cos(atan(2/3)).h
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