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Can this triangle be solved? 1 angle, 1side, and the sum of the other two sides known
- Thread starter Rycon9
- Start date
You could use the cosine law.
[FONT=KaTeX_Main][FONT=KaTeX_Math]a^[/FONT]2 =[FONT=KaTeX_Math]b^[/FONT] 2 +[FONT=KaTeX_Math]c^[/FONT] 2 −2[FONT=KaTeX_Math]b[/FONT][FONT=KaTeX_Math]c[/FONT](cos[FONT=KaTeX_Math]x[/FONT])[/FONT]
And this is painful to simplify and solve for a but it's doable.
If I have calculated correctly:
a = (2701-(2340*cos(71)))/(90-52cos(71)) ≈(2701-761.829)/(73.0705) ≈ (1939.171)/(73.071) ≈26.54
c = 45 - a ≈ 45 -26.54 ≈ 18.5 unit lengths.
[FONT=KaTeX_Main][FONT=KaTeX_Math]a^[/FONT]2 =[FONT=KaTeX_Math]b^[/FONT] 2 +[FONT=KaTeX_Math]c^[/FONT] 2 −2[FONT=KaTeX_Math]b[/FONT][FONT=KaTeX_Math]c[/FONT](cos[FONT=KaTeX_Math]x[/FONT])[/FONT]
And this is painful to simplify and solve for a but it's doable.
If I have calculated correctly:
a = (2701-(2340*cos(71)))/(90-52cos(71)) ≈(2701-761.829)/(73.0705) ≈ (1939.171)/(73.071) ≈26.54
c = 45 - a ≈ 45 -26.54 ≈ 18.5 unit lengths.
Last edited:
What are your thoughts? What have you tried? Please re-read the Read Before Postinghttps://www.freemathhelp.com/forum/threads/41537-Read-Before-Posting!! thread that's stickied at the top of each subforum, and comply with the rules found within. Specifically, please share with us any and all work you've done on this problem, even the parts you know for sure are wrong. Thank you.
Just in case you might be stuck at the very beginning, and so haven't shown any work because you have none to show, here's a big hint to push you in the right direction: Does the law of cosineshttps://www.mathsisfun.com/algebra/trig-cosine-law.html ring any bells? If you've been assigned this exercise, you ought to have learned about this recently. Another big hint: Yes, this exercise is completely solvable, and you will get a numerical value for c when all is said and done.
Just in case you might be stuck at the very beginning, and so haven't shown any work because you have none to show, here's a big hint to push you in the right direction: Does the law of cosineshttps://www.mathsisfun.com/algebra/trig-cosine-law.html ring any bells? If you've been assigned this exercise, you ought to have learned about this recently. Another big hint: Yes, this exercise is completely solvable, and you will get a numerical value for c when all is said and done.
There are 6 unknowns. We are given four equations, three explicitly and one implicitly.
\(\displaystyle A = 71.\)
\(\displaystyle b = 26.\)
\(\displaystyle a + c = 45 \implies c = 45 - a.\)
\(\displaystyle A + B + C = 180 \implies C = 180 - A - B = 109 - B.\)
Now consider the law of sines. Let sin(A) = sin(71) = p, which is a number not an unknown. Let q = sin(109) and r = cos(109). They are also numbers. Let s = sin(B), which is an unknown.
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(B)}{b} \implies \dfrac{p}{a} = \dfrac{s}{26} \implies a = \dfrac{26p}{s}.\)
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(C)}{c} = \dfrac{sin(109 - B)}{45 - a} \implies \dfrac{p}{a}= \dfrac{sin(109 - B)}{45 - a} \implies\)
\(\displaystyle a * sin(109 - B) = 45p - 45a \implies a\{sin(109 - B) + 45\} = 45p \implies a = \dfrac{45p}{sin(109 - B) + 45}.\)
Now we can use some identities.
\(\displaystyle sin(109 - B) = q * cos(B) - rs = q * \sqrt{1 - s^2} - rs.\)
\(\displaystyle \therefore \dfrac{s}{26} = a = \dfrac{45p}{q\sqrt{1 - s^2} - rs + 45} \implies 1170p = qs\sqrt{1 - s^2} - rs^2 + 45s \implies\)
\(\displaystyle 1170p + rs^2 - 45s = qs\sqrt{1 - s^2} \implies (1170p + rs^2 - 45s)^2 = q^2s^2 - q^2s^4.\)
If that quartic has a solution, then such a triangle exists. Of course there may be as many as four such triangles.
\(\displaystyle A = 71.\)
\(\displaystyle b = 26.\)
\(\displaystyle a + c = 45 \implies c = 45 - a.\)
\(\displaystyle A + B + C = 180 \implies C = 180 - A - B = 109 - B.\)
Now consider the law of sines. Let sin(A) = sin(71) = p, which is a number not an unknown. Let q = sin(109) and r = cos(109). They are also numbers. Let s = sin(B), which is an unknown.
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(B)}{b} \implies \dfrac{p}{a} = \dfrac{s}{26} \implies a = \dfrac{26p}{s}.\)
\(\displaystyle \dfrac{sin(A)}{a} = \dfrac{sin(C)}{c} = \dfrac{sin(109 - B)}{45 - a} \implies \dfrac{p}{a}= \dfrac{sin(109 - B)}{45 - a} \implies\)
\(\displaystyle a * sin(109 - B) = 45p - 45a \implies a\{sin(109 - B) + 45\} = 45p \implies a = \dfrac{45p}{sin(109 - B) + 45}.\)
Now we can use some identities.
\(\displaystyle sin(109 - B) = q * cos(B) - rs = q * \sqrt{1 - s^2} - rs.\)
\(\displaystyle \therefore \dfrac{s}{26} = a = \dfrac{45p}{q\sqrt{1 - s^2} - rs + 45} \implies 1170p = qs\sqrt{1 - s^2} - rs^2 + 45s \implies\)
\(\displaystyle 1170p + rs^2 - 45s = qs\sqrt{1 - s^2} \implies (1170p + rs^2 - 45s)^2 = q^2s^2 - q^2s^4.\)
If that quartic has a solution, then such a triangle exists. Of course there may be as many as four such triangles.