Trigonometry! #96

Let the angles be $x- d, x$ and $x + d$ for some $d \gt 0$ . Now since the three angles must sum to $180^{\circ}$ , we know that $x = 60^{\circ}$ . The other two angles then sum to $120^{\circ}$ , so for ease of calculation rename the largest angle $y$ , which makes the smallest angle $120^{\circ} - y$ .

Now the largest angle will be opposite the side length $10$ and the smallest angle will be opposite the shortest side. Letting this shortest side have length $L$ , the Sine Law gives us that

$\dfrac{\sin(120^{\circ} - y)}{L} = \dfrac{\sin(60^{\circ})}{9} = \dfrac{\sin(y)}{10}.$

From the second equality we have that

$\sin(y) = \dfrac{10\sin(60^{\circ})}{9} = \dfrac{5\sqrt{3}}{9}$ ,

and so $\cos(y) = \pm \sqrt{1 - \sin^{2}(y)} = \pm\sqrt{1 - \frac{75}{81}} = \pm \dfrac{\sqrt{6}}{9}.$

From the first equality we have that

$L = \dfrac{9\sin(120^{\circ} - y)}{\sin(60^{\circ})} = 6\sqrt{3}(\sin(120^{\circ})\cos(y) - \cos(120^{\circ})\sin(y)) =$

$6\sqrt{3}(\dfrac{\sqrt{3}}{2}*(\pm (\dfrac{\sqrt{6}}{9})) + \dfrac{1}{2}*\dfrac{5\sqrt{3}}{9}) = \pm\sqrt{6} + 5.$

Thus the two possible lengths of the third side are $5 - \sqrt{6}$ and $5 + \sqrt{6}$ , the serial numbers of which are 1 and 4, respectively.

Thus the desired sum is $1 + 4 = \boxed{5}$ .

For a = 10 and b = 9 opposite to angle A and B respectively,

Cos C = (10^2 + 9^2 - c^2)/ [2(10)(9)] = (181 - c^2)/ 180

Sin A/ 10 = Sin B/ 9 = Sin C/ c {Less safe} OR

Cos A = (9^2 + c^2 - 10^2)/ [2(9)(c)] and Cos B = (10^2 + c^2 - 9^2)/ [2(10)(c)]

For C, B, A:

14.20683095, 60, 105.793169;

31.17883554, 63.72786107, 85.09330339;

29.92643487, 63.89611886, 86.17744627;

45.79316905, 60, 74.20683095.

Using sine rule, 180 d - Asin x ought to be there for obtuse A; using cosine rule, we are safe from obtuse error. Differences:

45.7931690482639, 45.7931690482640;

32.5490255281136, 21.3654423278050;

33.9696839960459, 22.2813274080656;

14.2068309517360, 14.2068309517360.

Therefore 5 - Sqrt (6) and 5 + Sqrt (6) are the two answers suitable for A.P. angles.

1 + 4 = 5

Note: In this question, I tried by substitutions rather than to solve in order.