Trigonometry homework help question

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in triangle PQR, PQ = a, QR = a + d, RP = a + 2d, and angle PQR is more than 120 degree., where a and d are positive constants. show that

2a/3 < d < a.


i tried to think, but i really dont know how to start to solve this question. hope for some clue.

thank you
 

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AKG
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denian said:
in triangle PQR, PQ = a, QR = a + d, RP = a + 2d, and angle PQR is more than 120 degree., where a and d are positive constants. show that

2a/3 < d < a.


i tried to think, but i really dont know how to start to solve this question. hope for some clue.

thank you
Start with a picture, even if it doesn't help in the end, it's nice to have something to look at, you never know how it might inspire you. Now, we want to show:
1) d < a
2) d > 2a/3

The first part is easy. Assume [tex]d \geq a[/tex]. Let [tex]d = a + \epsilon[/tex], where [tex]\epsilon \geq 0[/tex], [tex]\epsilon \in \lR[/tex].

[tex]PQ + QR\ = 3a + \epsilon[/tex]
[tex]PR = 3a + 2\epsilon[/tex]

Therefore:

[tex]PR \geq PQ + QR[/tex]. Now, it should be obvious to you (especially if you try to draw it out with a ruler) that such a triangle is impossible. Our contradiction (that our triangle forms an impossible triangle) implies that our assumption is wrong (the assumption that [tex]d \geq a[/tex]), therefore, d is indeed less than a.

2) Now, we know that the [tex]\angle PQR = q[/tex] satisfies:

[tex]q > 120^o[/tex]

Therefore:

[tex]\cos q < -0.5[/tex]

Now, set up an equation using the Cosine Law:

[tex](a + 2d)^2 = a^2 + (a + d)^2 - 2a(a + d)\cos q[/tex]
[tex]\frac{a^2 - 2ad - 3d^2}{2a(a + d)} = \cos q[/tex]
[tex]\frac{a^2 - 2ad - 3d^2}{2a(a + d)} < -0.5[/tex]
[tex]a^2 - 2ad - 3d^2 < -a^2 - ad[/tex]
[tex]2a^2 - ad - 3d^2 < 0[/tex]
[tex](2a - 3d)(a + d) < 0[/tex]

Therefore, either:

[tex]2a - 3d < 0[/tex] AND [tex]a + d > 0[/tex], OR
[tex]2a - 3d > 0[/tex] AND [tex]a + d < 0[/tex]

It should be clearly impossible for "a + d < 0" to be true, therefore, we must meet the first set of conditions. Looking at "2a - 3d < 0", we get:

[tex]2a < 3d[/tex]
[tex]d > 2a/3[/tex]

This proves 2). Finally, we have it that 2a/3 < d < a. Q.E.D.
 
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Thank You!
 

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