Triangle area sum
Four DIFFERENT triangles have sides 4 and 6 and one of each of their's angels is 30 degrees. Given these measurements, there are 4 possible triangles that can be created with areas A, B, C, and D. Of these 4, A, B, and C have the greatest areas, Find A+B+C.
The answer is 26.88.
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3 solutions
First, we can get rid of the simple triangle with the 30 between the 4 and 6 because it has an area of 4
6
1/2*sin30= 6, so we will put that on the side and double check it later to see if it's one of the three greatest.
Next, when the 30 is across from the 6, we use law of sines so 6/sin(30)=4/sin x thus Sin(x)=1/3, using a calculator, we find x=160 or 19.4712 deg (we have two answers due to the identity Sin(x)=sin(180-x)), however, if the angle is 160 deg, we have two angles of 160 and 30 deg which add up to 190, making the next angle impossible because triangles can only have 180 deg. Thus two angles are 30 and 19.4712 and the remaining angle is 130.5288. Using 1/2absin(c) we find the area of this triangle to be 9.12.
Now this is where the real hard work comes in. If the 30 is across from the 4, 4/(.5)=6/Sin(x) thus Sinx=3/4 therefore. X=131.4096 or 48.5903, BOTH of these angles work. This is the reason why there are 4 possible triangles. This allows the remaining angle to be either 18.5904 or 101.4097. Therefore, the using 1/2absin(c), we find the area of the triangle to be either 3.825 or 11.763.
Remember, this problem asks for the 3 largest areas. We have areas 11.73, 9.12, 6, and 3.83. Out of these, the three largest are 11.73, 9.12, and 6. Thus 11.73+9.12+6=26.88
Consider a triangle △ A B C with A B = 4 and C = 6 . Now, let's move the angle to form three possible triangles:
The first one is ∠ A B C = 3 0 ° . Here, a simple application of sines law will give us the area, because the known sides are adjacent to the angle: A = 2 4 × 6 × sin ( 3 0 ° ) = 6
The second one is ∠ B A C = 3 0 ° . This time we will use cosines law. Let A C = x , so: B C 2 = A B 2 + A C 2 − 2 × A B × A C × cos ( 3 0 ° ) 6 2 = 4 2 + x 2 − 2 × 4 × x × cos ( 3 0 ° ) We get a quadratic equation: x 2 − 4 3 x − 2 0 = 0 This equation only has one positive solution, and it's, by the quadratic formula: x = 2 3 + 4 2 . With this, let's do the same than in the first triangle with sines law: B = 2 4 × x × sin ( 3 0 ° ) = 2 4 × ( 2 3 + 4 2 ) × sin ( 3 0 ° ) = 2 3 + 4 2 ≈ 9 . 1 2
The third one is ∠ A C B = 3 0 ° . Let's do the same: A B 2 = A C 2 + B C 2 − 2 × A C × B C × cos ( 3 0 ° ) 4 2 = x 2 + 6 2 − 2 × 6 × x × cos ( 3 0 ° ) x 2 − 6 3 x + 2 0 = 0 This time we have two positive solutions, that means that here we have two possible triangles. So, x = 3 3 ± 7 And the two possible areas are: C = 2 6 ( 3 3 + 7 ) sin ( 3 0 ° ) = 2 9 3 + 3 7 ≈ 1 1 . 7 6 And the second one: D = 2 6 ( 3 3 − 7 ) sin ( 3 0 ° ) = 2 9 3 − 3 7 ≈ 3 . 8 2
We have that A , B and C are the greatest areas. So, A + B + C = 6 + 9 . 1 2 + 1 1 . 7 6 = 2 6 . 8 8
I don't know why I couldn't post the solution directly.
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Nice job, this solution is much better than mine
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I just used Cosines Law to obtain exact values. Your solution is also nice, and you can put spaces between paragraphs putting two lines between each one.
That should be 11.76*, not 11.73. Also, how do you format it so there are spaces between paragraphs. It looks really ugly with no spaces.
Trevor, the other problem, post it again. It's still a good problem, but first, fix it.
Alright, I will, but I'll reword it and put in different numbers. Also, when you solved for the area of the hexagon with trisected angles, did you use law of sines a ton, and did you need a calculator? I'm trying to do it by hand and I'm kinda stuck.
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I just did it in the brute force way, computing the coordinates of the vertices of the hexagon, and then used the formula for finding the area of a polygon, given the coordinates of its vertices. It's a straightforward process, but very tedious, even with a calculator.
Trevor, given the changes, can you add your solution to this problem? Thanks.
L e t A B C b e t h e Δ . S i n 3 0 = 0 . 5 . W e w i l l u s e S i n L a w . a = 4 , b = 6 C = 3 0 o . a r e a = 2 1 ∗ a ∗ b ∗ S i n 3 0 = 1 2 ∗ . 5 . A = 3 0 o . a r e a = 2 1 ∗ a ∗ b ∗ S i n ( 3 0 + A r c S i n ( 4 6 ∗ S i n 3 0 ) ) = 1 2 ∗ . 9 8 0 2 4 . A = 3 0 o . a r e a = 2 1 ∗ a ∗ b ∗ S i n ( 3 0 + 1 8 0 − A r c S i n ( 4 6 ∗ S i n 3 0 ) ) = 1 2 ∗ . 3 1 8 8 . B = 3 0 o . a r e a = 2 1 ∗ a ∗ b ∗ S i n 3 0 = 1 2 ∗ . ( 3 0 + A r c S i n ( 6 4 ∗ S i n 3 0 ) ) = 1 2 ∗ . 7 6 0 1 . S o r e q u i r e d a r e a = 1 2 ∗ ( . 5 + . 9 8 0 2 4 + . 7 6 0 1 ) = 2 6 . 8 8 4 1 . Note:- 1] The blue area omitted is smallest. N o t e : − 2 ] S i n C = S i n ( A + B ) . ∴ S i n A a = S i n B b = S i n C c = S i n ( A + B ) c . B u t i f A = 3 0 , B = A r c S i n ( a b ∗ S i n A ) . ∴ S i n C = S i n ( 3 0 + A r c S i n ( a b ∗ S i n A ) ) S i m i l a r l y i f B = 3 0 . N o t e : − 3 ] A r c S i n X h a s t w o v a l u e s . A r c S i n X = α o r A r c S i n X = β , a n d α + β = 1 8 0 . ∴ When the only given angle is opposite the shorter given side, 2 triangles are formed i f α = β . However the obtuse angled triangle formed will have the least area as seen above.
SORRY I have used ABC for the triangle, same letters used in the problem to mean some thing else. I am too lazy to change the whole solution! Please pardon me.