Question

find f.
right triangle with one leg 11√2 mm, angles 60° and 30°, and the other leg labeled f
write your answer in simplest radical form.
blank millimeters
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Explanation:

Step1: Identify trigonometric ratio

In the right triangle, we can use the tangent function. For the \(30^\circ\) angle, \(\tan(30^\circ)=\frac{\text{opposite}}{\text{adjacent}}\). Wait, no, let's check the angles. The angle at the top is \(60^\circ\), the angle at the bottom is \(30^\circ\), and the right angle. The side opposite \(60^\circ\) is \(f\), and the side opposite \(30^\circ\) is \(11\sqrt{2}\) mm? Wait, no, let's label the triangle. Let's see: the right angle is between the side of length \(11\sqrt{2}\) and \(f\). So the angle of \(60^\circ\) is at the top, so the side adjacent to \(60^\circ\) is \(11\sqrt{2}\), and the side opposite \(60^\circ\) is \(f\). So \(\tan(60^\circ)=\frac{f}{11\sqrt{2}}\). Since \(\tan(60^\circ)=\sqrt{3}\), we have \(\sqrt{3}=\frac{f}{11\sqrt{2}}\).

Step2: Solve for \(f\)

Multiply both sides by \(11\sqrt{2}\): \(f = 11\sqrt{2}\times\sqrt{3}\). Using the property \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\), we get \(f = 11\sqrt{6}\). Wait, no, wait. Wait, maybe I mixed up the angles. Let's check again. The angle at the bottom is \(30^\circ\), so the side opposite \(30^\circ\) is \(11\sqrt{2}\), and the side adjacent to \(30^\circ\) is \(f\). So \(\tan(30^\circ)=\frac{11\sqrt{2}}{f}\). Since \(\tan(30^\circ)=\frac{1}{\sqrt{3}}\), then \(\frac{1}{\sqrt{3}}=\frac{11\sqrt{2}}{f}\). Cross - multiply: \(f = 11\sqrt{2}\times\sqrt{3}=11\sqrt{6}\)? Wait, no, that can't be. Wait, maybe using cotangent. Alternatively, since it's a 30 - 60 - 90 triangle, the ratios are \(1:\sqrt{3}:2\) for the sides opposite 30, 60, 90. Wait, the side opposite 30° is \(11\sqrt{2}\), so the side opposite 60° (which is \(f\)) should be \((11\sqrt{2})\times\sqrt{3}=11\sqrt{6}\)? Wait, no, let's do it with tangent correctly. Let's define the angles: the right angle is \(C\), angle \(A = 60^\circ\), angle \(B = 30^\circ\). Side \(AC = 11\sqrt{2}\) (adjacent to angle \(A\)), side \(BC = f\) (opposite to angle \(A\)). So \(\tan(A)=\tan(60^\circ)=\frac{BC}{AC}=\frac{f}{11\sqrt{2}}\). Since \(\tan(60^\circ)=\sqrt{3}\), then \(f = 11\sqrt{2}\times\sqrt{3}=11\sqrt{6}\). Wait, but let's check with 30 - 60 - 90 ratios. In a 30 - 60 - 90 triangle, the sides are in the ratio \(x:x\sqrt{3}:2x\), where \(x\) is the side opposite 30°, \(x\sqrt{3}\) opposite 60°, and \(2x\) hypotenuse. Here, the side opposite 30° is \(11\sqrt{2}\), so the side opposite 60° (which is \(f\)) should be \(x\sqrt{3}=(11\sqrt{2})\sqrt{3}=11\sqrt{6}\). Yes, that's correct.

Answer:

\(11\sqrt{6}\)