Question

1. a right triangle with hypotenuse ( 18sqrt{2} ), one angle ( 45^circ ), legs ( m ) and ( n ). 8) a right triangle with one leg ( sqrt{5} ), one angle ( 60^circ ), hypotenuse ( u ), leg ( v ). 9) a right triangle with one leg ( 8sqrt{3} ), one angle ( 60^circ ), leg ( y ), hypotenuse ( x ). 10) a right triangle with one leg ( \frac{13}{2} ), one angle ( 30^circ ), leg ( y ), hypotenuse ( x ).

Explanation:

Problem 7:

Step1: Identify triangle type

It's a 45 - 45 - 90 triangle (right - angled with one angle 45°), so legs are equal (\(m = n\)) and hypotenuse \(= m\sqrt{2}\).
Given hypotenuse \(= 18\sqrt{2}\), so \(m\sqrt{2}=18\sqrt{2}\), divide both sides by \(\sqrt{2}\), we get \(m = 18\), and since \(m = n\), \(n = 18\).

Step1: Identify triangle type

It's a 30 - 60 - 90 triangle (right - angled with one angle 60°). In a 30 - 60 - 90 triangle, if the shorter leg (opposite 30°) is \(a\), longer leg (opposite 60°) is \(a\sqrt{3}\), hypotenuse is \(2a\). Here, the leg opposite 30° is \(\sqrt{5}\)? Wait, no: the right angle and 60° angle, so the leg adjacent to 60° is \(\sqrt{5}\) (shorter leg? Wait, 60° angle: the side opposite 30° is shorter. Wait, angle at the top is 60°, so the right - angled triangle has angles 90°, 60°, 30°. So the side with length \(\sqrt{5}\) is adjacent to 60°, so it's the shorter leg (opposite 30°)? Wait, no: in a 30 - 60 - 90 triangle, the sides are in ratio \(1:\sqrt{3}:2\), where 1 is opposite 30°, \(\sqrt{3}\) opposite 60°, 2 opposite 90°. So if the side adjacent to 60° (opposite 30°) is \(a=\sqrt{5}\), then:

• Longer leg \(v=a\sqrt{3}=\sqrt{5}\times\sqrt{3}=\sqrt{15}\)? Wait, no, wait the angle is 60°, so the side with \(\sqrt{5}\) is the leg opposite 30°? Wait, no, the right angle, 60° angle, so the third angle is 30°. So the side labeled \(\sqrt{5}\) is adjacent to 60°, so it's the leg opposite 30°, so length \(a = \sqrt{5}\). Then:
• Hypotenuse \(u = 2a=2\sqrt{5}\)
• Longer leg \(v=a\sqrt{3}=\sqrt{5}\times\sqrt{3}=\sqrt{15}\)? Wait, no, wait I made a mistake. Wait, the side with \(\sqrt{5}\) is adjacent to 60°, so it's the leg opposite 30°, so the leg opposite 60° is \(v\), so \(v=\sqrt{5}\times\sqrt{3}=\sqrt{15}\), and hypotenuse \(u = 2\times\sqrt{5}\)? Wait, no, let's use trigonometry. \(\cos(60^{\circ})=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{\sqrt{5}}{u}\), and \(\cos(60^{\circ})=\frac{1}{2}\), so \(\frac{\sqrt{5}}{u}=\frac{1}{2}\), so \(u = 2\sqrt{5}\). And \(\tan(60^{\circ})=\frac{v}{\sqrt{5}}\), \(\tan(60^{\circ})=\sqrt{3}\), so \(v=\sqrt{5}\times\sqrt{3}=\sqrt{15}\).

Step1: Identify triangle type

30 - 60 - 90 triangle. The side with \(8\sqrt{3}\) is opposite 60°, so it's the longer leg (\(a\sqrt{3}\), where \(a\) is shorter leg). So \(a\sqrt{3}=8\sqrt{3}\), so \(a = 8\) (shorter leg, which is \(y\), opposite 30°). Then hypotenuse \(x = 2a=16\).

Answer:

\(m = 18\), \(n = 18\)

Problem 8: