So, here I am again, about to share yet another question I have encountered at Yahoo! Answers, although this time, I haven’t posted my answer on the site, because it took me quite a while to solve this thing. After all, I had to rely on the things I know to answer this, provided that I wasn’t exposed to problems of this type by that time.
Also, I happened to scroll down and accidentally had a glance at the other Top Contributors’ answers. I assure you this was unintentional, but, nevertheless, this was one of the factors that made me refuse to answer. As much as possible, I don’t want to see others’ answers when I haven’t answered yet. The other factor, of course, was the fact that I was baffled at the beginning that I didn’t have enough time to catch up.
Anyway, I think I have solved the problem. I saw this thing in one of our good Contributors’ answer: k/sinK
While this may seem a bit trivial, this served as a vital clue that aided me as I attempted to solve this math puzzle. Allow me to show you the whole question to break the ice.
Question: Letting the sides of an oblique triangle be k, l and m, k + l + m = 150 cm. If K = 42°, and L = 120°, what are the lengths of k, l and m?
With this said, I can safely assume that we are dealing with an oblique triangle, since M can only be 180° – 120° – 42° = 18°; no right angles. I don’t know, but law of sine and cosine always come to my mind when I see oblique triangles, which led me into thinking that law of sine might be involved in solving the problem. I don’t really know whether the k/sinK thing or my thoughts came first, but, whatever.
The next thing I did was to draw a triangle that describes the problem, although that was a bit unnecessary. Well, for my case, I was in a need of a diagram so as not to be lost in the problem, so yeah.
Afterwards, I proceeded into thinking whether the law of sine is indeed involved in the problem. While I found that it is truly essential in the solution in the end, let me show you what I did, step by step. It may appear a bit clumsy, though.
Assuming that the law of sine is indeed involved, it says that:
k/sinK = l/sinL = m/sinM
Which leads us to having:
k/sin42° = l/sin120° = m/sin18°
However, I knew that no matter how I manipulate this relationship, I won’t go any nearer to finding k, l and m. Also, I have tried to use substitution among the variables, like k = (150 – m – l), but I knew this was futile, considering that I don’t know k, l or m. Anyway, I reserved the relationships I’ve found, made through a little derivation just in case I’ll need them.
(k)sin120°/sin42° = l
(m)sin120° / sin18° = l
(l)sin42° / sin120° = k
(m)sin42° / sin18° = k
(k)sin18° / sin42° = m
(l)sin18° / sin120° = m
Immediately after writing these down, or probably before, I knew that I need to have a bunch of relationships to be taken into consideration so I can solve for k, l and m. That’s right, a system of equations! I already have one equation; k + l + m = 150. I then picked some of the relationships I’ve found while deriving to have the second and third equations. I ended up with this. From this point, I realized that the law of sine is indeed essential in this problem.
k + l + m = 150 ⇐ Eq. 1
(m)sin42° / sin18° + (k)sin120°/sin42° + (l)sin18° / sin120° = 150 ⇐ Eq. 2
(l)sin42° / sin120° = k ⇐ Eq. 3
I restricted my work to four decimal places, since I’m handling non-terminating decimals, thanks to sine.
k + l + m = 150
0.3878m + 0.2318k + 0.0639l = 26.8606
(l)sin42° = (k)sin120°
Gaussian Elimination isn’t new to me, and it wasn’t that hard to find k, l and m now. The idea of this method is to have two system of equations, with two variables each to lessen the work.
k + l + m = 150
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.2318(k + l + m = 150)
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.2318k – 0.2318l – 0.2318m = -34.77
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.1679l + 0.156m = -7.9094 (Eq. 1)
—-
k + l + m = 150
(l)sin42° = (k)sin120°
—-
-sin120°(k + l + m = 150)
(l)sin42° – (k)sin120° = 0
—-
-(k)sin120° – (l)sin120° – (m)sin120° = 129.9038
(l)sin42° – (k)sin120° = 0
—-
1.5352l + 0.8660m = 129.9038 (Eq. 2)
Now we have those two system of equations, we can now find l and m with much ease.
-0.1679l + 0.156m = -7.9094
1.5352l + 0.8660m = 129.9038
—-
0.156(1.5352l + 0.8660m = 129.9038)
-0.8660(-0.1679l + 0.156m = -7.9094)
—-
0.2395l = 20.2650
0.1459l = 6.8495
—-
0.3849l = 27.1145
l = 70.45 cm
Finding m:
-0.1679(70.45) + 0.156m = -7.9094
-11.8286 + 0.156m = -7.9094
0.156m = -7.9094 + 11.8286
m = 25.12 cm
Hence, k can only be:
k + l + m = 150
k = 150 – m – l
k = 150 – 25.12 – 70.45
k = 54.43 cm
Well, my answer is that k = 54.43 cm, l = 70.45 cm, and m = 25.12 cm. These dimensions form a real triangle, since it satisfies the condition that the sum of the two sides must be greater than the third. To prove my answer further, I tried to solve for the angles back using the answers I’ve got. If solving for the angles given three sides, you use the law of cosine to find first the largest angle.
l² = m² + k² – 2mk cosL
2mk cosL = m² + k² – l²
L = arccos([m² + k² - l²]/2mk)
L = arccos([25.12² + 54.43² - 70.45²]/2(25.12)(54.43))
L = 120°
Use the law of sine to find any of the two remaining angles.
k/sinK = l/sinL
sinK = ksinL/l
K = arcsin(ksinL/l)
K = arcsin([54.43]sin120°/70.45)
K = 42°
Naturally, M will be 18°. Hey, I think we got it. We got back to the angles we started with. But if you can see any mistakes in my solution, even a slight miscalculation, please do tell me, for I don’t want to mislead somebody. Anyway, I hope you appreciated the efforts exerted at this problem by someone who was not informed, although now enlightened, about the fact that you can actually solve for the sides of an oblique triangle, given the perimeter and two angles. Hope this helps!
Also, I happened to scroll down and accidentally had a glance at the other Top Contributors’ answers. I assure you this was unintentional, but, nevertheless, this was one of the factors that made me refuse to answer. As much as possible, I don’t want to see others’ answers when I haven’t answered yet. The other factor, of course, was the fact that I was baffled at the beginning that I didn’t have enough time to catch up.
Anyway, I think I have solved the problem. I saw this thing in one of our good Contributors’ answer: k/sinK
While this may seem a bit trivial, this served as a vital clue that aided me as I attempted to solve this math puzzle. Allow me to show you the whole question to break the ice.
Question: Letting the sides of an oblique triangle be k, l and m, k + l + m = 150 cm. If K = 42°, and L = 120°, what are the lengths of k, l and m?
With this said, I can safely assume that we are dealing with an oblique triangle, since M can only be 180° – 120° – 42° = 18°; no right angles. I don’t know, but law of sine and cosine always come to my mind when I see oblique triangles, which led me into thinking that law of sine might be involved in solving the problem. I don’t really know whether the k/sinK thing or my thoughts came first, but, whatever.
The next thing I did was to draw a triangle that describes the problem, although that was a bit unnecessary. Well, for my case, I was in a need of a diagram so as not to be lost in the problem, so yeah.
Afterwards, I proceeded into thinking whether the law of sine is indeed involved in the problem. While I found that it is truly essential in the solution in the end, let me show you what I did, step by step. It may appear a bit clumsy, though.
Assuming that the law of sine is indeed involved, it says that:
k/sinK = l/sinL = m/sinM
Which leads us to having:
k/sin42° = l/sin120° = m/sin18°
However, I knew that no matter how I manipulate this relationship, I won’t go any nearer to finding k, l and m. Also, I have tried to use substitution among the variables, like k = (150 – m – l), but I knew this was futile, considering that I don’t know k, l or m. Anyway, I reserved the relationships I’ve found, made through a little derivation just in case I’ll need them.
(k)sin120°/sin42° = l
(m)sin120° / sin18° = l
(l)sin42° / sin120° = k
(m)sin42° / sin18° = k
(k)sin18° / sin42° = m
(l)sin18° / sin120° = m
Immediately after writing these down, or probably before, I knew that I need to have a bunch of relationships to be taken into consideration so I can solve for k, l and m. That’s right, a system of equations! I already have one equation; k + l + m = 150. I then picked some of the relationships I’ve found while deriving to have the second and third equations. I ended up with this. From this point, I realized that the law of sine is indeed essential in this problem.
k + l + m = 150 ⇐ Eq. 1
(m)sin42° / sin18° + (k)sin120°/sin42° + (l)sin18° / sin120° = 150 ⇐ Eq. 2
(l)sin42° / sin120° = k ⇐ Eq. 3
I restricted my work to four decimal places, since I’m handling non-terminating decimals, thanks to sine.
k + l + m = 150
0.3878m + 0.2318k + 0.0639l = 26.8606
(l)sin42° = (k)sin120°
Gaussian Elimination isn’t new to me, and it wasn’t that hard to find k, l and m now. The idea of this method is to have two system of equations, with two variables each to lessen the work.
k + l + m = 150
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.2318(k + l + m = 150)
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.2318k – 0.2318l – 0.2318m = -34.77
0.3878m + 0.2318k + 0.0639l = 26.8606
—-
-0.1679l + 0.156m = -7.9094 (Eq. 1)
—-
k + l + m = 150
(l)sin42° = (k)sin120°
—-
-sin120°(k + l + m = 150)
(l)sin42° – (k)sin120° = 0
—-
-(k)sin120° – (l)sin120° – (m)sin120° = 129.9038
(l)sin42° – (k)sin120° = 0
—-
1.5352l + 0.8660m = 129.9038 (Eq. 2)
Now we have those two system of equations, we can now find l and m with much ease.
-0.1679l + 0.156m = -7.9094
1.5352l + 0.8660m = 129.9038
—-
0.156(1.5352l + 0.8660m = 129.9038)
-0.8660(-0.1679l + 0.156m = -7.9094)
—-
0.2395l = 20.2650
0.1459l = 6.8495
—-
0.3849l = 27.1145
l = 70.45 cm
Finding m:
-0.1679(70.45) + 0.156m = -7.9094
-11.8286 + 0.156m = -7.9094
0.156m = -7.9094 + 11.8286
m = 25.12 cm
Hence, k can only be:
k + l + m = 150
k = 150 – m – l
k = 150 – 25.12 – 70.45
k = 54.43 cm
Well, my answer is that k = 54.43 cm, l = 70.45 cm, and m = 25.12 cm. These dimensions form a real triangle, since it satisfies the condition that the sum of the two sides must be greater than the third. To prove my answer further, I tried to solve for the angles back using the answers I’ve got. If solving for the angles given three sides, you use the law of cosine to find first the largest angle.
l² = m² + k² – 2mk cosL
2mk cosL = m² + k² – l²
L = arccos([m² + k² - l²]/2mk)
L = arccos([25.12² + 54.43² - 70.45²]/2(25.12)(54.43))
L = 120°
Use the law of sine to find any of the two remaining angles.
k/sinK = l/sinL
sinK = ksinL/l
K = arcsin(ksinL/l)
K = arcsin([54.43]sin120°/70.45)
K = 42°
Naturally, M will be 18°. Hey, I think we got it. We got back to the angles we started with. But if you can see any mistakes in my solution, even a slight miscalculation, please do tell me, for I don’t want to mislead somebody. Anyway, I hope you appreciated the efforts exerted at this problem by someone who was not informed, although now enlightened, about the fact that you can actually solve for the sides of an oblique triangle, given the perimeter and two angles. Hope this helps!
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