Stuck on trigonometry question

Soldato
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Completely stuck on this trigonometry question:
2U4RhxB.png


Added some labels.....
sWTZfSy.png

So, obviously side "b" is easy enough, 8.544.

And I know that
sin(B) / b = sin(A) /a = sin(C) / c

So for the B/b angle/side:
0.866 / 8.544 = 0.101


But, I'm totally stumped at what to do next.


I could make it into some right angled triangles:
hec1w65.png

But not sure how that helps me, as neither triagle has enough information to use the basic SOH-CAH-TOA rules.

Hope someone can give me some pointers, thanks :)
 
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Completely stuck on this trigonometry question:
2U4RhxB.png


Added some labels.....
sWTZfSy.png

So, obviously side "b" is easy enough, 8.544.

And I know that
sin(B) / b = sin(A) /a = sin(C) / c

So for the B/b angle/side:
0.866 / 8.544 = 0.101


But, I'm totally stumped at what to do next.


I could make it into some right angled triangles:
hec1w65.png

But not sure how that helps me, as neither triagle has enough information to use the basic SOH-CAH-TOA rules.

Hope someone can give me some pointers, thanks :)

Surely if you have a value for the square root of 73, you can calculate the height of the green line in the last example. Because you have the angle of 90° on the other side of the perpendicular to work with. You can then calculate what X+2 is from the left hand triange and subtract 2 to get X.
 
Soldato
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Yeh realised after I posted.

Can you substitute into cosine rule then expand, factorise and rearrange to get x ?

I'd give it a go if I had pen and paper at hand.
 
Soldato
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(x+2)^2 + (2x-3)^2 = 73

Expand, factorise and rearrange to get x.... unless I'm missing something. Don't need to do any sinA triangle stuff...?

I really should have said in my opening post, I'm not good at maths, not even at GCSE level. I just don't have the knowledge of how to do what you're suggesting :O
I figured my answer would need to involved trigonometry rules, but perhaps that's where I'm going wrong - it's just another area of basic maths I need to know.
 
Soldato
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You can substitute what you currently have as the lengths of the triangle into the standard cosine rule.. however you have to then manipulate to isolate x on its own... so like you said it depends on whether you know how to do that. Tbh there may be a much quicker way then I'm suggesting, my way would get messy with alot of x's to collect and then having to factorise.
 
Soldato
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Using @lemonkettaz cosine rule:
73 = (X+2)² + (2X - 3)² -2(X+2)(2X-3)Cos(60)
73 = X² + 4X + 4 + 4X² -12X +9 -2X² -4X +3X +6
73 = 3X² -9X +19
0 = 3X² -9X -54
0 = X² - 3X -18

x = 6
 
Soldato
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Using @lemonkettaz cosine rule:
73 = (X+2)² + (2X - 3)² -2(X+2)(2X-3)Cos(60)
73 = X² + 4X + 4 + 4X² -12X +9 -2X² -4X +3X +6
73 = 3X² -9X +19
0 = 3X² -9X -54
0 = X² - 3X -18

x = 6

Very interesting, thanks!
Going to take a while to digest that, but I'm sure that'll keep me busy! :)

Thanks everyone.

EDIT: I see I've probably labelled my images incorrectly....I stuck the known length as "b", but I see you've started with 73 as "a". I guess you want to start with the known value on one side of the equation, and go from there?
 
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I can't for the life of me understand why complicating the problem with unnecessary equations and factorisation. I gave the solution in post 2, using basic trigonometry.

The rest of this thread is like describing how it's possible to remove a screw with a molewrench. That's always possible, but who in their right mind would resort to that if they had a screwdriver?

Go back to basics. The beauty of trigonometry is that you only need the length of one side of a triangle plus an angle to work all the rest out.

The OP gave himself the chance of resolving the question when he added the green perpendicular line.
 
Soldato
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The OP gave himself the chance of resolving the question when he added the green perpendicular line.
Thanks. Not sure what I'd do next to calculate the green line height though - on the right hand triangle, I've only got 1 side length, and the 90 degree angle....but I don't think you can use the 90' angle for the basic SOH-CAH-TOA rules?
 
Soldato
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I see I've probably labelled my images incorrectly....I stuck the known length as "b", but I see you've started with 73 as "a". I guess you want to start with the known value on one side of the equation, and go from there?

Ah, sorry. I rearranged it because of where the known angle is.
 
Soldato
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The beauty of trigonometry is that you only need the length of one side of a triangle plus an angle to work all the rest out.
That's not true. 1 angle + 1 side is not enough information.

Here's a rough sketch, the arrowed angles are identical and the red sides opposite them are identical lengths but clearly the other sides are different so you can't calculate the lengths with just 2 bits of information.
FufJlZb.png
 
Soldato
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I can't for the life of me understand why complicating the problem with unnecessary equations and factorisation. I gave the solution in post 2, using basic trigonometry.

The rest of this thread is like describing how it's possible to remove a screw with a molewrench. That's always possible, but who in their right mind would resort to that if they had a screwdriver?

Go back to basics. The beauty of trigonometry is that you only need the length of one side of a triangle plus an angle to work all the rest out.

The OP gave himself the chance of resolving the question when he added the green perpendicular line.

But if your plan is to use the right hand of the two right angle triangles formed by the green line, surely the issue is that you don't have an angle in that triangle (except the right angle of course)? You know one side length, and nothing else. Pretty sure you need two pieces of information, not just one, to extrapolate all the data about a right angle triangle.
It's a hell of a long time since I've done this though, so tell me if I'm wrong.
 
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