blackops23
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(i) show that cos(theta) = [b(a+c)]/2ac
(ii) if a=b and c=2b find the area of Triangle AOC
No idea how to do part (i)...cosine rule?
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Need help with trig/triangle question - HARD QUESTION!!!!! (1 Viewer)
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(i) show that cos(theta) = [b(a+c)]/2ac
(ii) if a=b and c=2b find the area of Triangle AOC
No idea how to do part (i)...cosine rule?
blackops23
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= D = D I got it!For part 1 consider areas. That should be a great enough hint
(I like this question a lot - makes you think.)
once you know the method it was suprisingly easy....but if someone didn't tell me to use areas I would never have been able to figure it out...
blackops23
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For part (ii) I got
A= {[4(a^2)(c^2) - (b^2)(a+c)^2]/2} - is it correct?
EDIT: is there an easier way to get the answer??
I did...Area of AOC = (b/2)(sin(theta))(a+c)
cos(theta) = [b(a+c)/2ac]....then I drew a right angled triangle....hyptonuse = 2ac, adjacent = b(a+c), and opposite was calculated via pythagoras...and sin(theta) was just opposite/hypotenuse....
is there a quicker way?
A= {[4(a^2)(c^2) - (b^2)(a+c)^2]/2} - is it correct?
EDIT: is there an easier way to get the answer??
I did...Area of AOC = (b/2)(sin(theta))(a+c)
cos(theta) = [b(a+c)/2ac]....then I drew a right angled triangle....hyptonuse = 2ac, adjacent = b(a+c), and opposite was calculated via pythagoras...and sin(theta) was just opposite/hypotenuse....
is there a quicker way?
Last edited:
asianese
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Umm check my working:
}{2ac}\cdot \sin\theta\\=\frac{b(a+c)}{2}\sin\theta\\ $Using a right triangle with sides $ 2ac, b(a+c) $ and $ \sqrt{4a^2c^2-b^2(a+c)^2}, $ along with the cosine relationship in part i, $\\ \sin\theta = \frac{\sqrt{4a^2c^2-b^2(a+c)^2}}{2ac}\\ \therefore |AOC| = \frac{b(a+c)}{4ac}\cdot \sqrt{4a^2c^2-b^2(a+c)^2})
I can't see a quicker way at the moment...You could consider sine rules within the smaller triangles but you'd be left with other angles that you don't know of. I think the 'ugly' answer is the cleanest/simplest one we can get.
Others please suggest otherwise
I can't see a quicker way at the moment...You could consider sine rules within the smaller triangles but you'd be left with other angles that you don't know of. I think the 'ugly' answer is the cleanest/simplest one we can get.
Others please suggest otherwise
Last edited:
D94
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If you sub in a=b and c=2b into the equation for cosØ, you get cosØ = 3/4. By trigonometry, sinØ = (√7)/4, therefore, sin2Ø = (3√7)/8. Area AOC = 1/2 ac sin2Ø = 1/2 b 2b (3√7)/8 = b2 (3√7)/8.
(I suspect there should be more to it, but that's what I got)
(I suspect there should be more to it, but that's what I got)
asianese
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Ah yes - that's correct.
But some extra thoughts:
The area of AOC doesn't actually depend on the length of B, nor the existence of OB itself. Would it be 'more correct' (for the sake of being nit picky) to say that the area is
? lols
But some extra thoughts:
The area of AOC doesn't actually depend on the length of B, nor the existence of OB itself. Would it be 'more correct' (for the sake of being nit picky) to say that the area is
D94
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The question states: "if a=b and c=2b find the area of Triangle AOC".