Find x in the two triangles.

Okay, let's solve for x in both triangles. Triangle 6: We have a right triangle ATP with \(\angle P = 90^{\circ}\). Also, sides AP and PT are equal (both marked as x), so \(\triangle ATP\) is an isosceles right triangle. This means that \(\angle A = \angle T = 45^{\circ}\). Side AT has length 30. We can use trigonometry to find the value of x. Let's use the tangent function: $$\tan(A) = \frac{PT}{AP} = \frac{x}{x} = 1$$ Since \(\angle A = 45^{\circ}\), we can use the cosine function to find x: $$\cos(45^{\circ}) = \frac{AP}{AT} = \frac{x}{30}$$ We know that \(\cos(45^{\circ}) = \frac{\sqrt{2}}{2}\), so: $$\frac{\sqrt{2}}{2} = \frac{x}{30}$$ Multiply both sides by 30: $$x = 30 \cdot \frac{\sqrt{2}}{2} = 15\sqrt{2}$$ So, for triangle 6: $$ x = 15\sqrt{2} $$ Triangle 7: We have a right triangle ESR with \(\angle R = 90^{\circ}\) and \(\angle S = 60^{\circ}\). Therefore, \(\angle E = 30^{\circ}\). We are given SR = 9, and ES = x. We can use trigonometry to find x. Since we know the side opposite to angle E (SR) and we want to find the hypotenuse (ES), we can use the sine function: $$\sin(E) = \frac{SR}{ES}$$ $$\sin(30^{\circ}) = \frac{9}{x}$$ We know that \(\sin(30^{\circ}) = \frac{1}{2}\), so: $$\frac{1}{2} = \frac{9}{x}$$ Cross-multiply: $$x = 2 \cdot 9 = 18$$ So, for triangle 7: $$ x = 18 $$ Triangle 6: $$x = 15\sqrt{2}$$ Triangle 7: $$x = 18$$