Question

1. find the value of x.

enter your answer.

Explanation:

Step1: Identify triangle type

Triangle \(ABC\) is right - angled at \(A\), and \(\angle C = 45^{\circ}\), so it's a 45 - 45 - 90 triangle. In a 45 - 45 - 90 triangle, the legs are equal, and the hypotenuse \(c\) and leg \(a\) are related by \(c=a\sqrt{2}\), or \(a = \frac{c}{\sqrt{2}}\). Here, hypotenuse \(BC = 6.4\), and we want to find leg \(AB=x\).

Step2: Apply 45 - 45 - 90 triangle formula

For a 45 - 45 - 90 triangle, the length of each leg \(l\) is given by \(l=\frac{\text{hypotenuse}}{\sqrt{2}}\). So \(x=\frac{6.4}{\sqrt{2}}\). Rationalizing the denominator, we multiply numerator and denominator by \(\sqrt{2}\): \(x = \frac{6.4\sqrt{2}}{2}=3.2\sqrt{2}\approx3.2\times1.414 = 4.5248\)? Wait, no, wait. Wait, in a 45 - 45 - 90 triangle, the legs are equal, and hypotenuse \(=\) leg \(\times\sqrt{2}\). Wait, maybe I mixed up. Let's re - examine the triangle. Angle at \(A\) is right angle, angle at \(C\) is \(45^{\circ}\), so angle at \(B\) is also \(45^{\circ}\) (since sum of angles in a triangle is \(180^{\circ}\), \(180 - 90 - 45=45\)). So triangle \(ABC\) is an isosceles right - triangle with \(AB = AC\) (legs) and \(BC\) (hypotenuse). So hypotenuse \(BC = 6.4\), and leg \(AB=x\). The formula for hypotenuse \(h\) in terms of leg \(l\) is \(h = l\sqrt{2}\), so \(l=\frac{h}{\sqrt{2}}\). So \(x=\frac{6.4}{\sqrt{2}}\). Let's calculate this: \(\frac{6.4}{\sqrt{2}}\approx\frac{6.4}{1.4142}\approx4.525\)? Wait, no, wait, maybe I made a mistake. Wait, if we consider the sides: in a 45 - 45 - 90 triangle, the legs are equal, and hypotenuse is leg \(\times\sqrt{2}\). So if we let the leg be \(x\), then hypotenuse \(=x\sqrt{2}\). We know hypotenuse is \(6.4\), so \(x\sqrt{2}=6.4\), so \(x=\frac{6.4}{\sqrt{2}}\). But maybe we can also use trigonometry. \(\sin(45^{\circ})=\frac{AB}{BC}\), so \(AB = BC\times\sin(45^{\circ})\). Since \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071\), then \(AB = 6.4\times\frac{\sqrt{2}}{2}=3.2\sqrt{2}\approx4.525\). Wait, but maybe the triangle is labeled differently? Wait, angle at \(C\) is \(45^{\circ}\), right angle at \(A\), so side \(AB\) is opposite angle \(C\), side \(AC\) is adjacent to angle \(C\), and \(BC\) is hypotenuse. So \(\sin(C)=\frac{AB}{BC}\), so \(AB = BC\times\sin(45^{\circ})\). \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\approx0.7071\), so \(AB = 6.4\times\frac{\sqrt{2}}{2}=3.2\sqrt{2}\approx4.53\) (or if we use exact form, \(3.2\sqrt{2}\) or \(\frac{6.4}{\sqrt{2}}\) which simplifies to \(3.2\sqrt{2}\)). But maybe there's a simpler way. Wait, if we rationalize \(\frac{6.4}{\sqrt{2}}\), multiply numerator and denominator by \(\sqrt{2}\): \(\frac{6.4\sqrt{2}}{2}=3.2\sqrt{2}\approx4.52\). Alternatively, maybe the problem is using the fact that in a 45 - 45 - 90 triangle, the legs are equal to hypotenuse divided by \(\sqrt{2}\), but maybe we can also note that \(6.4\div\sqrt{2}\approx4.53\). Wait, but let's check with decimal approximation. \(\sqrt{2}\approx1.4142\), so \(6.4\div1.4142\approx4.525\), which is approximately \(4.53\) or if we use exact value, \(3.2\sqrt{2}\). But maybe the problem expects an exact form or a decimal. Wait, maybe I made a mistake in the triangle labeling. Wait, the triangle has vertices \(B\), \(C\), \(A\) with right angle at \(A\), so \(AB\) and \(AC\) are legs, \(BC\) is hypotenuse. Angle at \(C\) is \(45^{\circ}\), so triangle is isosceles right - triangle, so \(AB = AC\). So hypotenuse \(BC = 6.4\), so \(AB=\frac{6.4}{\sqrt{2}} = 3.2\sqrt{2}\approx4.53\). So the value of \(x\) (which is \(AB\)) is \(\frac{6.4}{\sqrt{2}}\) or \(3.2\sqrt{2}\) or approximately…

Answer:

\(3.2\sqrt{2}\) (or approximately \(4.53\))