# Trigonometry homework help question

1. May 21, 2004

### denian

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

2. May 21, 2004

### AKG

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 $$d \geq a$$. Let $$d = a + \epsilon$$, where $$\epsilon \geq 0$$, $$\epsilon \in \lR$$.

$$PQ + QR\ = 3a + \epsilon$$
$$PR = 3a + 2\epsilon$$

Therefore:

$$PR \geq PQ + QR$$. 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 $$d \geq a$$), therefore, d is indeed less than a.

2) Now, we know that the $$\angle PQR = q$$ satisfies:

$$q > 120^o$$

Therefore:

$$\cos q < -0.5$$

Now, set up an equation using the Cosine Law:

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

Therefore, either:

$$2a - 3d < 0$$ AND $$a + d > 0$$, OR
$$2a - 3d > 0$$ AND $$a + d < 0$$

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:

$$2a < 3d$$
$$d > 2a/3$$

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

3. May 29, 2004

Thank You!