Question

consider right triangle △ghi below. which expressions represent the length of side \\(\overline{hi}\\)? choose 2 answers:

Explanation:

To solve for the length of \( \overline{HI} \) in right triangle \( \triangle GHI \) (right - angled at \( I \)):

Step 1: Recall trigonometric ratios

In a right - triangle, the three main trigonometric ratios are:

• Sine of an angle: \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine of an angle: \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent of an angle: \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \)

In \( \triangle GHI \), \( \angle H = 75^{\circ} \), \( \angle I=90^{\circ} \), \( GI = 8 \), \( GH=8.3 \)

Step 2: Analyze the sides with respect to \( \angle H \)

• For \( \angle H = 75^{\circ} \):
• The side opposite to \( \angle H \) is \( GI = 8 \)
• The side adjacent to \( \angle H \) is \( HI \)
• The hypotenuse is \( GH = 8.3 \)

Step 3: Use the cosine and sine (or Pythagorean theorem) to find expressions for \( HI \)

• Using cosine ratio:

The cosine of an angle \( \theta \) in a right - triangle is defined as \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \theta = 75^{\circ} \), adjacent side is \( HI \) and hypotenuse is \( GH = 8.3 \). So, \( \cos(75^{\circ})=\frac{HI}{GH} \), and \( HI = 8.3\times\cos(75^{\circ}) \)

• Using sine ratio (with respect to the other non - right angle):

The sum of angles in a triangle is \( 180^{\circ} \). So, \( \angle G=180^{\circ}-\angle H-\angle I=180 - 75-90 = 15^{\circ} \)
For \( \angle G = 15^{\circ} \), the opposite side is \( HI \) and the hypotenuse is \( GH = 8.3 \). So, \( \sin(15^{\circ})=\frac{HI}{GH} \), and \( HI = 8.3\times\sin(15^{\circ}) \)

• Using Pythagorean theorem:

In a right - triangle, \( (GI)^{2}+(HI)^{2}=(GH)^{2} \). We know that \( GI = 8 \) and \( GH = 8.3 \). So, \( HI=\sqrt{(GH)^{2}-(GI)^{2}}=\sqrt{8.3^{2}-8^{2}} \)

• Using tangent ratio:

\( \tan(75^{\circ})=\frac{GI}{HI} \), so \( HI=\frac{GI}{\tan(75^{\circ})}=\frac{8}{\tan(75^{\circ})} \)

If we consider the most straightforward trigonometric ratio applications:

1. From \( \cos(75^{\circ})=\frac{HI}{8.3} \), we get \( HI = 8.3\cos(75^{\circ}) \)
2. From \( \tan(75^{\circ})=\frac{8}{HI} \), we get \( HI=\frac{8}{\tan(75^{\circ})} \) (or from \( \sin(15^{\circ})=\frac{HI}{8.3} \), \( HI = 8.3\sin(15^{\circ}) \))

Suppose the options (not fully provided but based on the analysis) could be:

• Option 1: \( 8.3\cos(75^{\circ}) \)
• Option 2: \( \frac{8}{\tan(75^{\circ})} \) (or \( 8.3\sin(15^{\circ}) \))

If we assume the options are in the form of these expressions, the two correct expressions for the length of \( \overline{HI} \) are \( 8.3\cos(75^{\circ}) \) and \( \frac{8}{\tan(75^{\circ})} \) (or \( 8.3\sin(15^{\circ}) \))

Answer:

To solve for the length of \( \overline{HI} \) in right triangle \( \triangle GHI \) (right - angled at \( I \)):

Step 1: Recall trigonometric ratios

In a right - triangle, the three main trigonometric ratios are:

• Sine of an angle: \( \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}} \)
• Cosine of an angle: \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \)
• Tangent of an angle: \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \)

In \( \triangle GHI \), \( \angle H = 75^{\circ} \), \( \angle I=90^{\circ} \), \( GI = 8 \), \( GH=8.3 \)

Step 2: Analyze the sides with respect to \( \angle H \)

• For \( \angle H = 75^{\circ} \):
• The side opposite to \( \angle H \) is \( GI = 8 \)
• The side adjacent to \( \angle H \) is \( HI \)
• The hypotenuse is \( GH = 8.3 \)

Step 3: Use the cosine and sine (or Pythagorean theorem) to find expressions for \( HI \)

• Using cosine ratio:

The cosine of an angle \( \theta \) in a right - triangle is defined as \( \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} \). For \( \theta = 75^{\circ} \), adjacent side is \( HI \) and hypotenuse is \( GH = 8.3 \). So, \( \cos(75^{\circ})=\frac{HI}{GH} \), and \( HI = 8.3\times\cos(75^{\circ}) \)

• Using sine ratio (with respect to the other non - right angle):

The sum of angles in a triangle is \( 180^{\circ} \). So, \( \angle G=180^{\circ}-\angle H-\angle I=180 - 75-90 = 15^{\circ} \)
For \( \angle G = 15^{\circ} \), the opposite side is \( HI \) and the hypotenuse is \( GH = 8.3 \). So, \( \sin(15^{\circ})=\frac{HI}{GH} \), and \( HI = 8.3\times\sin(15^{\circ}) \)

• Using Pythagorean theorem:

In a right - triangle, \( (GI)^{2}+(HI)^{2}=(GH)^{2} \). We know that \( GI = 8 \) and \( GH = 8.3 \). So, \( HI=\sqrt{(GH)^{2}-(GI)^{2}}=\sqrt{8.3^{2}-8^{2}} \)

• Using tangent ratio:

\( \tan(75^{\circ})=\frac{GI}{HI} \), so \( HI=\frac{GI}{\tan(75^{\circ})}=\frac{8}{\tan(75^{\circ})} \)

If we consider the most straightforward trigonometric ratio applications:

1. From \( \cos(75^{\circ})=\frac{HI}{8.3} \), we get \( HI = 8.3\cos(75^{\circ}) \)
2. From \( \tan(75^{\circ})=\frac{8}{HI} \), we get \( HI=\frac{8}{\tan(75^{\circ})} \) (or from \( \sin(15^{\circ})=\frac{HI}{8.3} \), \( HI = 8.3\sin(15^{\circ}) \))

Suppose the options (not fully provided but based on the analysis) could be:

• Option 1: \( 8.3\cos(75^{\circ}) \)
• Option 2: \( \frac{8}{\tan(75^{\circ})} \) (or \( 8.3\sin(15^{\circ}) \))

If we assume the options are in the form of these expressions, the two correct expressions for the length of \( \overline{HI} \) are \( 8.3\cos(75^{\circ}) \) and \( \frac{8}{\tan(75^{\circ})} \) (or \( 8.3\sin(15^{\circ}) \))