Question

3.
∠g=\underline{28°}
∠h=\underline{}
∠i=\underline{}
(there is a triangle ghi with angle at g being 28°, and sides gh and gi marked as equal? wait, no, looking at the diagram: triangle ghi, with a mark on gh and a mark on gi? wait, the diagram shows triangle ghi, with angle at g is 28°, and side gh has a tick mark, side gi has a tick mark? wait, no, maybe gh and hi? wait, the original diagram: point g, h, i. angle at g is 28°, side gh has a tick, side hi has a tick? wait, the users diagram: h and i, with x° at h. so triangle ghi, with angle g = 28°, and sides gh and hi are equal? wait, the tick marks: one on gh, one on hi. so its an isosceles triangle with gh = hi? wait, no, maybe gh = gi? wait, the ocr text: the problem is to find angles h and i, with angle g given as 28°, and triangle with two sides marked (gh and hi? or gh and gi? the diagram as per user: triangle ghi, angle at g is 28°, side gh has a tick, side hi has a tick? wait, the users input: the image shows triangle ghi, with angle at g is 28°, and side gh (from g to h) has a tick, side hi (from h to i) has a tick? wait, no, maybe gh and gi. wait, the key is: in triangle ghi, angle g is 28°, and two sides are equal (marked with ticks), so its an isosceles triangle. so we need to find angles h and i. so the ocr text is the problem to find ∠h and ∠i, given ∠g=28° and the triangle is isosceles (with gh=hi? or gh=gi? wait, the diagram: h is at the bottom left, i at bottom right, g at top. so angle at g is 28°, and sides gh and hi are equal? wait, no, maybe gh and gi. wait, regardless, the problem is about triangle angles, isosceles triangle, so sum of angles in triangle is 180°, so if two sides are equal, two angles are equal. so the ocr text is the problem to calculate angles h and i in triangle ghi with ∠g=28° and its isosceles (gh=hi or gh=gi or hi=gi).

Explanation:

Step1: Identify the triangle type

The triangle \( \triangle GHI \) has two sides marked as equal (the marks on \( GH \) and \( GI \)? Wait, no, looking at the diagram, \( GH \) and \( HI \)? Wait, no, the marks: one on \( GH \) and one on \( GI \)? Wait, no, the diagram shows \( GH \) with one mark and \( HI \) with one mark? Wait, no, actually, the two sides with marks are \( GH \) and \( GI \)? Wait, no, the angle at \( G \) is \( 28^\circ \), and two sides are equal (the marks on \( GH \) and \( GI \)? Wait, no, the triangle has two sides equal, so it's an isosceles triangle. Wait, the marks: one on \( GH \) and one on \( HI \)? Wait, no, the diagram: \( H \) has \( x^\circ \), \( I \) has a mark, \( GH \) has a mark. So actually, \( GH = GI \)? Wait, no, the marks: \( GH \) and \( HI \) have marks? Wait, no, the problem: the triangle has two sides equal, so the base angles are equal? Wait, no, the vertex angle is \( \angle G = 28^\circ \), and the two equal sides are \( GH \) and \( GI \)? Wait, no, the marks: \( GH \) and \( HI \) are marked as equal? Wait, no, the diagram: \( H \) is a vertex, \( I \) is a vertex, \( G \) is the top. The sides \( GH \) and \( HI \) are marked? Wait, no, the marks: \( GH \) has one mark, \( HI \) has one mark? Wait, no, the correct interpretation: in triangle \( GHI \), \( GH = GI \) (the marks), so it's isosceles with \( GH = GI \), so the base angles are \( \angle H \) and \( \angle I \)? Wait, no, that can't be. Wait, no, the angle at \( G \) is \( 28^\circ \), and the two equal sides are \( GH \) and \( HI \), so \( \angle G \) is the vertex angle, and \( \angle H \) and \( \angle I \) are the base angles? Wait, no, let's re-examine.

Wait, the diagram: \( G \) is the top, \( H \) and \( I \) are the base vertices. The side \( GH \) has a mark, \( HI \) has a mark? No, the marks: \( GH \) and \( GI \) have marks? Wait, no, the problem says \( \angle G = 28^\circ \), and two sides are equal (the marks on \( GH \) and \( GI \)). Wait, no, the correct approach: in an isosceles triangle, the sum of angles is \( 180^\circ \). So if \( \angle G = 28^\circ \), and the two equal sides are \( GH \) and \( GI \), then the base angles \( \angle H \) and \( \angle I \) are equal? Wait, no, that would mean \( \angle H = \angle I \), but the side \( HI \) is the base. Wait, no, maybe the equal sides are \( GH \) and \( HI \), so \( \angle G = \angle I \)? Wait, no, the angle at \( G \) is \( 28^\circ \), so that can't be. Wait, I think I made a mistake. Let's start over.

The triangle \( \triangle GHI \) has \( \angle G = 28^\circ \), and two sides are equal (marked as equal: \( GH = HI \)? Wait, no, the marks: \( GH \) has one mark, \( HI \) has one mark? No, the diagram shows \( GH \) and \( GI \) with marks? Wait, no, the problem: the triangle has two equal sides, so it's isosceles. The angle at \( G \) is \( 28^\circ \), so the other two angles ( \( \angle H \) and \( \angle I \)): wait, no, if \( GH = GI \), then \( \angle H = \angle I \). Wait, no, the sum of angles in a triangle is \( 180^\circ \). So \( \angle G + \angle H + \angle I = 180^\circ \). If \( \angle G = 28^\circ \), and \( \angle H = \angle I \) (since \( GH = GI \)), then \( 28 + 2x = 180 \), so \( 2x = 152 \), \( x = 76 \). Wait, but the diagram shows \( \angle H = x^\circ \), and \( \angle I \) has a mark, \( GH \) has a mark. Wait, maybe the equal sides are \( HI \) and \( GI \), so \( \angle G = \angle H \)? No, that doesn't make sense. Wait, the correct interpretation: the triangle has \( GH = HI \), so \( \angle G = \angle I = 28^\circ \), then \( \angle H = 180 - 28 - 28 = 124^\circ \). Wait, but the angle at \( H \) is \( x^\circ \), which is obtuse. Let's check the diagram again.

Looking at the diagram: \( H \) is a vertex with \( x^\circ \), \( I \) is a vertex with a mark, \( G \) is the top with \( 28^\circ \). The sides \( GH \) and \( HI \) are marked as equal (the marks). So \( GH = HI \), so the triangle is isosceles with \( GH = HI \), so the base angles are \( \angle G \) and \( \angle I \), and the vertex angle is \( \angle H \). Wait, no, in an isosceles triangle, the angles opposite equal sides are equal. So if \( GH = HI \), then the angles opposite them are \( \angle I \) and \( \angle G \), respectively. So \( \angle I = \angle G = 28^\circ \). Then \( \angle H = 180 - 28 - 28 = 124^\circ \). So \( x = 124 \), \( \angle I = 28^\circ \), \( \angle G = 28^\circ \). Wait, but the problem says \( \angle G = 28^\circ \), so that matches. So:

Step1: Determine the triangle type

The triangle \( \triangle GHI \) has \( GH = HI \) (marked as equal), so it's isosceles with \( \angle G = \angle I \) (angles opposite equal sides).

Step2: Calculate \( \angle I \)

Since \( GH = HI \), \( \angle I = \angle G = 28^\circ \).

Step3: Calculate \( \angle H \)

Sum of angles in a triangle: \( \angle G + \angle H + \angle I = 180^\circ \)
Substitute \( \angle G = 28^\circ \) and \( \angle I = 28^\circ \):
\( 28 + x + 28 = 180 \)
\( x + 56 = 180 \)
\( x = 180 - 56 = 124 \)

So \( \angle G = 28^\circ \), \( \angle H = 124^\circ \), \( \angle I = 28^\circ \).

Answer:

\( \angle G = 28^\circ \), \( \angle H = 124^\circ \), \( \angle I = 28^\circ \)