Trig Word problem

[amathproblemthatneedsolve]

A student is talented in furniture maker and has decided to build a bike frame out of wood. The student likes this design they found online, but instead of curves on the frame, the student will use straight pieces of wood. As the student intends to use a wooden bike to demonstrate some woodworking skills and promote his business, they will place a triangular sign inside the frame. The sign must have an area of at least 800 cm2. the student has taken some measurements from the old steel road bike and will use these when they create his new wooden masterpiece.


1. Find other crucial dimensions.

(you are expected to demonstrate the use of the sine rule, the cosine rule and the area of a triangle rule.)

 

[Dr.Peterson]

I think you know by now that we will ask you to show some work, so we can tell where you need help.

What parts are you able to do? I see a good place to use the cosine rule to find a length; and a place where you can find another length that will give the desired (minimum) area. Give it a try and show us.

 

[amathproblemthatneedsolve]

Sorry, Dr. Peterson, my question was how to start the problem? once I know how to start it I should be able to work it out

 

[Harry_the_cat]

Redraw the smaller triangle on its own and mark in the sides and angles that you know. Can you find the missing side?

 

[Dr.Peterson]

Sorry, Dr. Peterson, my question was how to start the problem? once I know how to start it I should be able to work it out

In general, the way to start a problem is to look at it and think about what you can do. You don't need someone else to tell you exactly what to do. That is, your goal is to learn to solve problems on your own, so you must not become dependent on others to do this initial thinking for you. That's why I only gave a couple little hints, leaving you free to practice this important skill. But they were hints!

Harry's suggestion is what I hinted at when I said, "I see a good place to use the cosine rule to find a length".

 

[amathproblemthatneedsolve]

Redraw the smaller triangle on its own and mark in the sides and angles that you know. Can you find the missing side?

I tried it, please refer to the attachment as my LateX is bad. So angle "c" is 23.68

 

[Dr.Peterson]

You meant side c (the segment common to the two triangles), right?

Now look at the big triangle. You know an angle, a side, and the area. What can you do there?

 

[amathproblemthatneedsolve]

You meant side c (the segment common to the two triangles), right?

Now look at the big triangle. You know an angle, a side, and the area. What can you do there?

Yes. My angle is 70 degrees, side length is 23.68? and the area is 800[MATH]cm^2[/MATH]
I tried doing(I know 'tried doing something isn't very clear) something and I got (a) (Above the 70degrees angle line to the right) length of 71.9 and the other length to be 67.57004

 

[Dr.Peterson]

Did you forget to add the 15 cm to the 23.68?

 

[amathproblemthatneedsolve]

Yeah, whoops. SO if side a is 71.9 and side b is 30.68 (15+23.68) side c is 68.63686 (note a,b,c are just place holders, I'll draw a graph)

 

[Dr.Peterson]

Check your addition. 15+23.68 = 30.68?

And why 71.9? Isn't that the number you got using the wrong numbers?

 

[amathproblemthatneedsolve]

@Dr.Peterson , Yeah your right, I should have thought more before I posted. Do I need to work out the other angles also?

Solving for b

Solving for a-

 

[Dr.Peterson]

Your lengths look good now; what else you need to do depends on your interpretation of "other crucial dimensions". But since you haven't yet used the sine rule (I think), I imagine they want you to find the angles.

 

[amathproblemthatneedsolve]

Which formula would I need to use?( I'll look also)

 

[Dr.Peterson]

I think I gave you a strong hint ...

 

[amathproblemthatneedsolve]

Well, I didn't use sin,e I used cosine but here we go was this wrong?.

.

 

[Dr.Peterson]

You're using [MATH]\alpha[/MATH] for the angle at A, right? You were told that it is 70°, so that's encouraging.

The sine rule is easier to use here, when you know two sides and an angle, but what you did avoids possible issues with obtuse angles, so it's a good idea. And I get that way the same angles you got your way.

 

[amathproblemthatneedsolve]

You're using [MATH]\alpha[/MATH] for the angle at A, right? You were told that it is 70°, so that's encouraging.

The sine rule is easier to use here, when you know two sides and an angle, but what you did avoids possible issues with obtuse angles, so it's a good idea. And I get that way the same angles you got your way.

I believe I am. If you don't mind could I see your working using the sine rule? to compare it to mine?

 

[Dr.Peterson]

I think you did every step for the larger triangle differently than I did, so I'll do all of that (using what I believe are your labels).

First, we have [MATH]\alpha = 70^{\circ}[/MATH], b = 38.68, and area = 800. I used the formula [MATH]Area = \frac{bc}{2}\sin\alpha[/MATH], so

[MATH]800 = \frac{38.68}{2}c\sin70^{\circ}[/MATH], therefore [MATH]c = \frac{2\cdot800}{38.68\sin70^{\circ}} = 44.02[/MATH]​

To find your side a, I would use the cosine rule, [MATH]a^2 = b^2 + c^2 - 2bc\cos{\alpha}[/MATH]:

[MATH]a^2 = 38.68^2 + 44.02^2 - 2(38.68)(44.02)\cos{70^{\circ}} = 2269.2[/MATH]​

[MATH]a = \sqrt{2269.2} = 47.64[/MATH]​

To find your angle [MATH]\beta[/MATH], I would use [MATH]\frac{a}{\sin\alpha} = \frac{b}{\sin\beta}[/MATH]:

[MATH]\sin\beta = \frac{b\sin\alpha}{a} = \frac{38.68\sin70^{\circ}}{47.64} = 0.76396[/MATH]​

[MATH]\beta = \sin^{-1} 0.76296 = 49.73^{\circ}[/MATH]​

Different people in different places have different styles. Yours is very interesting.

 

[amathproblemthatneedsolve]

And to get the last angle you simply add 70+49.73 and deduct that from 180 to get 60.27degree. Thanks for your help!