$3$ . Given right triangle $G H I$ , with right angle at $H, G H=12.2$ , and $m \angle G=28^{\circ}$ , find the measures of the remaining sides to the nearest tenth.

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Grade 6

Find missing angles in special triangles

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3

. Given right triangle

G H I

, with right angle at

H, G H=12.2

, and

m \angle G=28^{\circ}

, find the measures of the remaining sides to the nearest tenth.

Use Trigonometric Ratios: To find the lengths of the remaining sides of the triangle, we can use trigonometric ratios. Since we have the length of one side ( $GH$ ) and the measure of one acute angle (angle $G$ ), we can use the sine and cosine functions to find the lengths of $HI$ (opposite side) and $GI$ (hypotenuse).
Find Length of HI: First, we'll find the length of HI using the sine function. The sine of an angle in a right triangle is equal to the length of the opposite side divided by the length of the hypotenuse. Since we don't have the hypotenuse yet, we'll rearrange the formula to solve for the opposite side: $\newline$ $\sin(G) = \frac{\text{opposite}}{HI}$ $\newline$ $HI = GH \times \sin(G)$
Calculate Length of GI: Now we'll plug in the values we know: $\newline$ $HI = 12.2 \times \sin(28^\circ)$ $\newline$ Using a calculator, we find that $\sin(28^\circ) \approx 0.4695$ . $\newline$ $HI = 12.2 \times 0.4695$
Calculate Length of GI: Now we'll plug in the values we know: $\newline$ $HI = 12.2 \times \sin(28^\circ)$ $\newline$ Using a calculator, we find that $\sin(28^\circ) \approx 0.4695$ . $\newline$ $HI = 12.2 \times 0.4695$ Calculating the length of HI: $\newline$ $HI \approx 12.2 \times 0.4695$ $\newline$ $HI \approx 5.7289$ $\newline$ To the nearest tenth, $HI \approx 5.7$ cm.
Calculate Length of GI: Now we'll plug in the values we know: $\newline$ $HI = 12.2 \times \sin(28^\circ)$ $\newline$ Using a calculator, we find that $\sin(28^\circ) \approx 0.4695$ . $\newline$ $HI = 12.2 \times 0.4695$ Calculating the length of HI: $\newline$ $HI \approx 12.2 \times 0.4695$ $\newline$ $HI \approx 5.7289$ $\newline$ To the nearest tenth, $HI \approx 5.7$ cm.Next, we'll find the length of GI using the cosine function. The cosine of an angle in a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse. Since GH is the adjacent side to angle G, we can write: $\newline$ $\cos(G) = \frac{GH}{GI}$ $\newline$ $GI = \frac{GH}{\cos(G)}$
Calculate Length of GI: Now we'll plug in the values we know: $\newline$ $HI = 12.2 \times \sin(28^\circ)$ $\newline$ Using a calculator, we find that $\sin(28^\circ) \approx 0.4695$ . $\newline$ $HI = 12.2 \times 0.4695$ Calculating the length of HI: $\newline$ $HI \approx 12.2 \times 0.4695$ $\newline$ $HI \approx 5.7289$ $\newline$ To the nearest tenth, $HI \approx 5.7$ cm.Next, we'll find the length of GI using the cosine function. The cosine of an angle in a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse. Since GH is the adjacent side to angle G, we can write: $\newline$ $\cos(G) = \frac{GH}{GI}$ $\newline$ $GI = \frac{GH}{\cos(G)}$ Plugging in the values we know: $\newline$ $GI = \frac{12.2}{\cos(28^\circ)}$ $\newline$ Using a calculator, we find that $\cos(28^\circ) \approx 0.8829$ . $\newline$ $\sin(28^\circ) \approx 0.4695$ $0$
Calculate Length of GI: Now we'll plug in the values we know: $\newline$ $HI = 12.2 \times \sin(28^\circ)$ $\newline$ Using a calculator, we find that $\sin(28^\circ) \approx 0.4695$ . $\newline$ $HI = 12.2 \times 0.4695$ Calculating the length of HI: $\newline$ $HI \approx 12.2 \times 0.4695$ $\newline$ $HI \approx 5.7289$ $\newline$ To the nearest tenth, $HI \approx 5.7$ cm.Next, we'll find the length of GI using the cosine function. The cosine of an angle in a right triangle is equal to the length of the adjacent side divided by the length of the hypotenuse. Since GH is the adjacent side to angle G, we can write: $\newline$ $\cos(G) = \frac{GH}{GI}$ $\newline$ $GI = \frac{GH}{\cos(G)}$ Plugging in the values we know: $\newline$ $GI = \frac{12.2}{\cos(28^\circ)}$ $\newline$ Using a calculator, we find that $\cos(28^\circ) \approx 0.8829$ . $\newline$ $\sin(28^\circ) \approx 0.4695$ $0$ Calculating the length of GI: $\newline$ $\sin(28^\circ) \approx 0.4695$ $1$ $\newline$ $\sin(28^\circ) \approx 0.4695$ $2$ $\newline$ To the nearest tenth, $\sin(28^\circ) \approx 0.4695$ $3$ cm.