Question

triangle 5: choose a number between 30-50 for the opposite side, and between 5-18 for theta. find the adjacent side and the hypotenuse.
adjacent:
hypotenuse:

Explanation:

Step1: Choose values for Opposite and Theta

Let's choose Opposite side \( o = 40 \) (between 30 - 50) and Theta \( \theta = 60^\circ \) (a common angle in right - triangles). We know that in a right - triangle, \( \sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}} \) and \( \cos\theta=\frac{\text{Adjacent}}{\text{Hypotenuse}} \), also \( \tan\theta=\frac{\text{Opposite}}{\text{Adjacent}} \)

Step2: Calculate Hypotenuse using sine

We know that \( \sin\theta=\frac{o}{h} \), so \( h = \frac{o}{\sin\theta} \). For \( \theta = 60^\circ \), \( \sin60^\circ=\frac{\sqrt{3}}{2}\approx0.866 \) and \( o = 40 \). Then \( h=\frac{40}{\sin60^\circ}=\frac{40}{\frac{\sqrt{3}}{2}}=\frac{80}{\sqrt{3}}\approx46.19 \) (we can also choose a simpler angle, say \( \theta = 30^\circ \), \( \sin30^\circ = 0.5 \), then \( h=\frac{40}{0.5}=80 \))

Step3: Calculate Adjacent using cosine

Using \( \cos\theta=\frac{a}{h} \), so \( a = h\cos\theta \). If we take \( \theta = 30^\circ \), \( \cos30^\circ=\frac{\sqrt{3}}{2}\approx0.866 \) and \( h = 80 \), then \( a=80\times\cos30^\circ = 80\times\frac{\sqrt{3}}{2}=40\sqrt{3}\approx69.28 \). Alternatively, using \( \tan\theta=\frac{o}{a} \), \( a=\frac{o}{\tan\theta} \). For \( \theta = 30^\circ \), \( \tan30^\circ=\frac{1}{\sqrt{3}}\approx0.577 \), \( a = \frac{40}{\tan30^\circ}=40\sqrt{3}\approx69.28 \)

Let's choose a simpler case where \( \theta = 45^\circ \), Opposite \( o = 40 \). Then \( \tan45^\circ = 1=\frac{o}{a}\), so \( a = 40 \). And \( \sin45^\circ=\frac{\sqrt{2}}{2}\approx0.707 \), \( h=\frac{40}{\sin45^\circ}=\frac{40}{\frac{\sqrt{2}}{2}} = 40\sqrt{2}\approx56.57 \)

Let's pick whole numbers for easier calculation. Let Opposite \( o = 30 \), Theta \( \theta = 30^\circ \)

Step4: Recalculate with whole numbers

• Hypotenuse: \( \sin30^\circ=\frac{1}{2}=\frac{30}{h}\Rightarrow h = 60 \)
• Adjacent: \( \cos30^\circ=\frac{\sqrt{3}}{2}=\frac{a}{60}\Rightarrow a = 60\times\frac{\sqrt{3}}{2}=30\sqrt{3}\approx51.96 \). Or using \( \tan30^\circ=\frac{30}{a}\Rightarrow a=\frac{30}{\tan30^\circ}=30\sqrt{3}\approx51.96 \)

If we choose Opposite \( o = 50 \), Theta \( \theta = 45^\circ \)

• Hypotenuse: \( \sin45^\circ=\frac{\sqrt{2}}{2}=\frac{50}{h}\Rightarrow h = 50\sqrt{2}\approx70.71 \)
• Adjacent: \( \tan45^\circ = 1=\frac{50}{a}\Rightarrow a = 50 \)

Let's take a more straightforward approach. Let's choose Opposite side \( o = 30 \), Theta \( \theta = 30^\circ \)

• Hypotenuse (\( h \)): Since \( \sin\theta=\frac{\text{Opposite}}{\text{Hypotenuse}} \), \( h=\frac{\text{Opposite}}{\sin\theta}=\frac{30}{\sin30^\circ}=\frac{30}{0.5} = 60 \)
• Adjacent (\( a \)): Since \( \cos\theta=\frac{\text{Adjacent}}{\text{Hypotenuse}} \), \( a = h\times\cos\theta=60\times\cos30^\circ=60\times\frac{\sqrt{3}}{2}=30\sqrt{3}\approx52 \) (or using \( \tan\theta=\frac{\text{Opposite}}{\text{Adjacent}} \), \( a=\frac{\text{Opposite}}{\tan\theta}=\frac{30}{\tan30^\circ}=30\sqrt{3}\approx52 \))

Another example: Let Opposite \( o = 40 \), Theta \( \theta = 53.13^\circ \) (since \( \sin53.13^\circ\approx0.8 \), \( \cos53.13^\circ\approx0.6 \))

• Hypotenuse: \( h=\frac{40}{\sin53.13^\circ}=\frac{40}{0.8}=50 \)
• Adjacent: \( a = h\times\cos53.13^\circ=50\times0.6 = 30 \) (here we get whole numbers, which is simpler)

Answer:

Let's take Opposite \( = 40 \), Theta \( = 53.13^\circ \) (approximate angle where \( \sin\theta = 0.8 \), \( \cos\theta = 0.6 \))

• Adjacent Side \( = 30 \)
• Hypotenuse \( = 50 \)

(Or other valid combinations based on the chosen Opposite side and angle. For example, if Opposite = 30, Theta = 30°, Adjacent≈52, Hypotenuse = 60)