Question

for side lengths of right triangles mining missing components of right triangles use the information given in the diagram of △gmi to find the missing measurements. round all answers to the nearest tenth.

Explanation:

To solve for the missing components of right triangle \( \triangle GHI \) (with \( \angle H = 90^\circ \), \( \angle I = 30^\circ \), and hypotenuse \( GI = 30 \)):

Step 1: Find \( m\angle G \)

In a right triangle, the sum of angles is \( 180^\circ \). Since \( \angle H = 90^\circ \) and \( \angle I = 30^\circ \):
\[
m\angle G = 180^\circ - 90^\circ - 30^\circ = 60^\circ
\]

Step 2: Find \( HI \) (adjacent to \( 30^\circ \))

In a \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) triangle, the side adjacent to \( 30^\circ \) (longer leg) is \( \frac{\sqrt{3}}{2} \times \text{hypotenuse} \). Here, hypotenuse \( GI = 30 \):
\[
HI = \frac{\sqrt{3}}{2} \times 30 = 15\sqrt{3} \approx 25.98
\]

Step 3: Find \( GH \) (opposite to \( 30^\circ \))

In a \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) triangle, the side opposite \( 30^\circ \) (shorter leg) is \( \frac{1}{2} \times \text{hypotenuse} \):
\[
GH = \frac{1}{2} \times 30 = 15
\]

Final Answers:

• \( m\angle G = \boldsymbol{60^\circ} \)
• \( HI \approx \boldsymbol{25.98} \) (or \( 15\sqrt{3} \))
• \( GH = \boldsymbol{15} \)

Answer:

To solve for the missing components of right triangle \( \triangle GHI \) (with \( \angle H = 90^\circ \), \( \angle I = 30^\circ \), and hypotenuse \( GI = 30 \)):

Step 1: Find \( m\angle G \)

In a right triangle, the sum of angles is \( 180^\circ \). Since \( \angle H = 90^\circ \) and \( \angle I = 30^\circ \):
\[
m\angle G = 180^\circ - 90^\circ - 30^\circ = 60^\circ
\]

Step 2: Find \( HI \) (adjacent to \( 30^\circ \))

In a \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) triangle, the side adjacent to \( 30^\circ \) (longer leg) is \( \frac{\sqrt{3}}{2} \times \text{hypotenuse} \). Here, hypotenuse \( GI = 30 \):
\[
HI = \frac{\sqrt{3}}{2} \times 30 = 15\sqrt{3} \approx 25.98
\]

Step 3: Find \( GH \) (opposite to \( 30^\circ \))

In a \( 30^\circ\text{-}60^\circ\text{-}90^\circ \) triangle, the side opposite \( 30^\circ \) (shorter leg) is \( \frac{1}{2} \times \text{hypotenuse} \):
\[
GH = \frac{1}{2} \times 30 = 15
\]

Final Answers:

• \( m\angle G = \boldsymbol{60^\circ} \)
• \( HI \approx \boldsymbol{25.98} \) (or \( 15\sqrt{3} \))
• \( GH = \boldsymbol{15} \)