Question 1189764: To solve triangle ABC(with angles A,B and C, sides a, b, and c, with side length la, lb, lc respectively)
(A) If A = 45°, la=7sqrt2, and lb=7 then there is one solution.
(B) If A=30°, la < Ib/2, then there is no solution.
(C) If A = 30°, la > Ib/2, then there are two solutions.
(D) All the above (A) (B) (C) are true.
(E) None of the above (A) (B) (C) is true.
Answer: D
Hey, I need help to solve this please, without a calculator, i solved the first and second one(by picking random values) but i cant solve the third one. How can i do it?
Found 3 solutions by math_tutor2020, ikleyn, MathTherapy:
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Just for clarification, something like lb is Lb with lowercase L. I'll use the uppercase version.
Angles: A, B, C
sides: a,b,c which are opposite A,B,C in that order
La = length of side a
Lb = length of side b
Lc = length of side c
Let's see if claim (A) is true
A = 45
La = 7*sqrt(2)
Lb = 7
Use the law of sines
sin(A)/La = sin(B)/Lb
Lb*sin(A) = La*sin(B)
7*sin(45) = 7*sqrt(2)*sin(B)
7*sqrt(2)/2 = 7*sqrt(2)*sin(B)
7/2 = 7sin(B) ................ the sqrt(2) terms cancel
sin(B) = (7/2)*(1/7)
sin(B) = 1/2
B = arcsin(1/2)
B = 30 or B = 180-30 = 150
If B = 30, then
C = 180-A-B
C = 180-45-30
C = 105
Showing that B = 30 is valid
If B = 150, then,
C = 180-A-B
C = 180-45-150
C = -15
Showing that B = 150 isn't possible.
Therefore, the only practical solution is B = 30
In other words, only one triangle is possible.
We've proven claim (A) to be true. This rules out claim (E)
Either the final answer is (A) or it's (D).
Let's check the next claim.
=======================================================
Claim (B) If A = 30 and La < Lb/2, then there is no solution
Let's say
La = 10
Lb = 30
There's nothing special about those values. I just picked two random values to make La < Lb/2 true.
Apply the law of sines to find B
sin(A)/La = sin(B)/Lb
sin(45)/10 = sin(B)/30
30*sin(45) = 10*sin(B)
30*sqrt(2)/2 = 10*sin(B)
15sqrt(2) = 10*sin(B)
sin(B) = 15*sqrt(2)/10
sin(B) = 1.5*sqrt(2)
sin(B) = 2.1213 approximately
We run into a problem because the largest sin(B) can get is 1.
There are no solutions to sin(B) = 2.1213
Therefore, no such triangle is possible with the properties that
A = 30
La = 10
Lb = 30
More generally, no such triangle can be formed where
A = 30
La < Lb/2
The conclusion here is that claim (B) is true.
Claims (A) and (B) being true immediately points to choice (D) being the final answer (aka all of the above).
=======================================================
We don't have to keep going at this point, but for the sake of completeness, let's check claim (C)
Claim: If A = 30 and La > Lb/2, then there are two solutions
Let's say
La = 30
Lb = 10
sin(A)/La = sin(B)/Lb
sin(30)/30 = sin(B)/10
10*sin(30) = 30*sin(B)
10*(1/2) = 30*sin(B)
5 = 30*sin(B)
sin(B) = 5/30
sin(B) = 1/6
The sin(B) being a value smaller than 1 indicates sin(B) = 1/6 is possible to solve.
Use the inverse sine function to find that the two possible B values are
B = 9.594 or B = 180-9.594 = 170.406
If B = 9.594, then,
C = 180-A-B
C = 180-45-9.594
C = 125.406
Showing that we have a valid angle B here
Or if B = 170.406, then,
C = 180-A-B
C = 180-45-170.406
C = -35.406
That's not good. We should have gotten a positive C value.
It turns out that La > Lb/2 isn't quite the full story.
Instead, it should be
Lb/2 < La < Lb
If that compound inequality is true and A = 30, then two triangles are possible.
On the other hand, if La > Lb, then no triangles are possible.
I recommend using a tool like GeoGebra to play around with possible constructions.
It's not clear if your teacher intends La < Lb or not. It appears your teacher is implying this (or else choice D isn't possible). I would ask for further clarification.
For more info, check out this page
https://math.libretexts.org/Bookshelves/Precalculus/Book%3A_Elementary_Trigonometry_(Beveridge)/04%3A_The_Law_of_Sines_and_The_Law_of_Cosines/4.02%3A_The_Law_of_Sines_-_the_ambiguous_case
To wrap things up, statements (A), (B), and (C) are true. This is why choice (D) is the final answer. Again this is assuming your teacher implies that La < Lb.
Answer by ikleyn(52800)  (Show Source):
You can put this solution on YOUR website! .
Hello, instead of posting this GIBBERISH, consider to hire somebody, who will decipher
your hieroglyphs and re-write your notes correctly after you.
Otherwise, I will REGULARLY DESTROY/delete your posts.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
To solve triangle ABC(with angles A,B and C, sides a, b, and c, with side length la, lb, lc respectively)
(A) If A = 45°, la=7sqrt2, and lb=7 then there is one solution.
(B) If A=30°, la < Ib/2, then there is no solution.
(C) If A = 30°, la > Ib/2, then there are two solutions.
(D) All the above (A) (B) (C) are true.
(E) None of the above (A) (B) (C) is true.
Answer: D
Hey, I need help to solve this please, without a calculator, i solved the first and second one(by picking random values) but i cant solve the third one. How can i do it?
For C), if la > lb, then ONLY 1 triangle is possible. However, if lb > la, then 2 are possible.
There needs to be some other CLUE/INDICATION stating whether la or lb is LONGER.
|
|
|
