10. The lengths of the sides of a triangle \( A B C \) are \( x-2, x \) and \( x+2 \). The largest angle is \( 120^{\circ} \). (a) Find the value of \( x \). (b) Show that the area of the triangle is \( \frac{15 \sqrt{3}}{4} \) (c) Find \( \sin A+\sin B+\sin C \), giving your answer in the form \( \frac{p \sqrt{q}}{r} \) where \( p, q, r \in \mathbb{Z} \).

Let's delve into the solution for the triangle with side lengths \( x-2, x, \) and \( x+2 \), where the largest angle \( C \) is given to be \( 120^\circ \). ### (a) Finding the value of \( x \) Since \( C \) is the largest angle opposite the longest side \( x+2 \), we can use the Law of Cosines: \[ (x+2)^2 = (x-2)^2 + x^2 - 2(x-2)x \cos(120^\circ) \] Since \( \cos(120^\circ) = -\frac{1}{2} \), we have: \[ (x+2)^2 = (x-2)^2 + x^2 + (x-2)x \] Expanding both sides: \[ x^2 + 4x + 4 = x^2 - 4x + 4 + x^2 + x^2 - 2x \] Combining like terms: \[ x^2 + 4x + 4 = 3x^2 - 6x + 4 \] \[ 0 = 2x^2 - 10x \] \[ 0 = 2x(x - 5) \] From this, we find \( x = 0 \) (not valid for sides of a triangle) or \( x = 5 \). ### (b) Area of the triangle Now, substituting \( x = 5 \): The side lengths become \( 3, 5, \) and \( 7 \). Using Heron's formula to calculate the area: 1. Calculate the semi-perimeter \( s = \frac{3 + 5 + 7}{2} = 7.5 \). 2. Area \( A = \sqrt{s(s-a)(s-b)(s-c)} \): \[ A = \sqrt{7.5(7.5-3)(7.5-5)(7.5-7)} \] \[ = \sqrt{7.5 \cdot 4.5 \cdot 2.5 \cdot 0.5} \] \[ = \sqrt{7.5 \times 4.5 \times 2.5 \times 0.5} = \sqrt{42.1875} = \frac{15\sqrt{3}}{4} \] ### (c) Finding \(\sin A + \sin B + \sin C\) Using the sine rule: We have the angle \( C = 120^\circ \), and we find \( \sin C = \frac{\sqrt{3}}{2} \). To find angles \( A \) and \( B \), we can find \( \sin A \) and \( \sin B \) using the side lengths \( 3, 5, 7 \): \[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] With \( c = 7, a = 3, b = 5 \): \[ \frac{3}{\sin A} = \frac{7}{\sin 120^\circ} \Rightarrow \sin A = \frac{3 \cdot \frac{\sqrt{3}}{2}}{7} = \frac{3\sqrt{3}}{14} \] \[ \frac{5}{\sin B} = \frac{7}{\sin 120^\circ} \Rightarrow \sin B = \frac{5 \cdot \frac{\sqrt{3}}{2}}{7} = \frac{5\sqrt{3}}{14} \] Now, summing them: \[ \sin A + \sin B + \sin C = \frac{3\sqrt{3}}{14} + \frac{5\sqrt{3}}{14} + \frac{\sqrt{3}}{2} \] Convert \(\frac{\sqrt{3}}{2}\) to a common denominator: \[ \sin A + \sin B + \sin C = \frac{3\sqrt{3} + 5\sqrt{3} + 7\sqrt{3}}{14} = \frac{15\sqrt{3}}{14} \] Thus, your values are \( p = 15, q = 3, r = 14 \).