easy trig problem?

[abeefaria]

I can't remember how to solve this, any help would be appreciated, thank you.

[cosine]

I can't remember how to solve this, any help would be appreciated, thank you.

Haha, it really reminds me of this funny site. Are you sure thats not a right triangle? If not, I'm not sure if there is enough information to solve the problem.

[Primarygun]

Haha, it really reminds me of this funny site. Are you sure thats not a right triangle? If not, I'm not sure if there is enough information to solve the problem.

yes, I agree with you.

I think the question lacks information for the finding.

The line joining from the tip of the triangle is not restricted to any point on the horizontal line.

[gagsrcool]

Hi,

 

You cannot solve the problem. So, the answer is D).

 

gagsrcool

[Klaynos]

Normally if it's not a right angled triangle they try and make them not look them to save confussion... :|

[doG]

Normally if it's not a right angled triangle they try and make them not look them to save confussion... :|

Even if it were marked a right angle there would not be enough information to solve the problem.

[entwined]

Even if it were marked a right angle there would not be enough information to solve the problem.

 

it looks to me like if it were a 90 degree angle, that side X would be the sine of a 21 degree angle (.3584) times the length of the hypotenuse (325.345) or 116.603.

 

That is, carried out to 3 places. No?

[doG]

it looks to me like if it were a 90 degree angle' date=' that side X would be the sine of a 21 degree angle (.3584) times the length of the hypotenuse (325.345) or 116.603.

 

That is, carried out to 3 places. No?[/quote']

Yes, on my screen the little 21° in the corner looks like 2r which didn't make any sense. You are correct though, if it is 21° we can use the Law of Sines to find the hypotenuse from the information given and the hypotenuse to find x.

[entwined]

Yes, on my screen the little 21° in the corner looks like 2r which didn't make any sense. You are correct though, if it is 21° we can use the Law of Sines to find the hypotenuse from the information given and the hypotenuse to find x.

 

I see what you mean if you are looking at the shaded diagram rather that the hand drawn one.

 

I cannot see how to solve this if the opposite angle is not 90 degrees.

[EvoN1020v]

According to my calculation, the answer is C.

 

First, I assume the corner angle to be 90 degrees. It has to be 90 degrees so in the small triangle, 44 + 90 = 134. That leaves 46 degrees for the other lone angle. To verify this theory, Add all the angles in the big triangle (21 + 23 + 46 + 90), thus it equals to 180.

 

Then, I find the length of X.

[math]sin44 = \frac{X}{167.84} = 116.5914609[/math]

 

Then to verify the length of X:

[math]c^{2} = a^{2} + b^{2}[/math]

[math]167.84^{2} = a^{2} + 116.5914609^{2}[/math]

[math]X = 120.7339921[/math]

 

As you see that the small triangle's angle is 90 and 44, the both length side-by-side 90 degrees should be almost the same. Because on a right angle triangle the 2 other degrees are 45 degrees. Thus, the opposite and adajecent sides to 45 degrees, are the same length. The point is, the small triangle should have its opposite and adajecent sides almost the same which is evidentally showed: 116.5914609 and 120.7339921.

 

To verify the lengths of the sides:

[math]c^{2} = a^{2} + b^{2}[/math]

[math]325.345^{2} - (183 + 120.7339921)^{2} = b^{2}[/math]

[math]=\sqrt{13,595.03107} = b[/math]

[math]X = 116.5977318[/math]

 

So my guess remains at C following my proofs.

(Now I need to go and do my own homework. )

[doG]

First, I assume the corner angle to be 90 degrees. It has to be 90 degrees so in the small triangle, 44 + 90 = 134.

Bad assumption. If it were 90° though:

 

sin23/183 = sin 136/h by the law of sines. This yields h=325.345. Now x=h*sin21=116.593, not 116.591 or 116.597 and certainly not c (120.74). Also, if x = 120.74 and we call the unknown angle y we can write sin21/120.74 = siny/325.345, thus y=74.94°, not 90°.

[EvoN1020v]

You're absolutely right doG. Also I checked the x = sin44(167.84) and it yielded 116.591409, compare to 116.5932205 for x = sin21(325.345).

 

Also, I checked the confirmation of the left small triangle using one of the trigonometric laws: [math]C^{2} = A^{2} + B^{2}-2ABsinC[/math].

 

[math]C^{2} = 183^{2} + 167.84^{2} - 2(183)(167.84)cos136[/math]

It yielded [math]A = 325.342[/math].

 

Regardless, I still think we can figure out the answer for X. But, I'm not exactly sure how. Maybe we just can't, because we need one more value. (i.e. angle).

[doG]

Regardless, I still think we can figure out the answer for X. But, I'm not exactly sure how. Maybe we just can't, because we need one more value. (i.e. angle).

Do the math for all 3 answer choices of X. You'll find one solution for X=183.2 and 2 solutions for X=120.74. I've given the acute answer for X=120.74 above but there is also an obtuse solution as well. There is no solution for X=96.58 (Can you figure out why?). That means that D is also incorrect since there are valid solutions for choices A & C.

[EvoN1020v]

I concur with DoG. I couldn't stop thinking about this problem for the last 2 days, so I just decided to actual draw the triangles and see how it makes out. I just divided the lengths by 100 and the numbers behaved as centrimeters. Also, I used a projector (spelling?) for the correct drawing of the angles.

 

In my conclusion, either A or C are correct.