Question

triangle abc and triangle efg are shown below. what is ( mangle g )? (\boxed{})°

Explanation:

Step1: Analyze triangle ABC

Triangle \( ABC \) is a right triangle with \( \angle B = 90^\circ \). The sum of angles in a triangle is \( 180^\circ \), so \( \angle A + \angle C + \angle B = 180^\circ \). Substituting the known values: \( (6x + 3) + (5x - 1) + 90 = 180 \).

Step2: Solve for x

Simplify the equation: \( 6x + 3 + 5x - 1 + 90 = 180 \)
Combine like terms: \( 11x + 92 = 180 \)
Subtract 92 from both sides: \( 11x = 180 - 92 = 88 \)
Divide by 11: \( x = \frac{88}{11} = 8 \).

Step3: Find \( \angle A \)

Substitute \( x = 8 \) into \( \angle A \): \( 6(8) + 3 = 48 + 3 = 51^\circ \).

Step4: Analyze triangle EFG

Triangles \( ABC \) and \( EFG \) are congruent (marked sides and right angles), so \( \angle G = \angle C \) or we can use angle correspondence. First, find \( \angle C \): \( 5x - 1 = 5(8) - 1 = 40 - 1 = 39^\circ \)? Wait, no—wait, in triangle \( ABC \), angles sum to 180. Wait, \( \angle A = 51^\circ \), \( \angle B = 90^\circ \), so \( \angle C = 180 - 90 - 51 = 39^\circ \)? Wait, no, wait: \( (6x + 3) + (5x - 1) = 90 \) (since right triangle, other two angles sum to 90). Oh! I made a mistake earlier. In a right triangle, the two acute angles sum to \( 90^\circ \), so \( (6x + 3) + (5x - 1) = 90 \). That's a better approach.

Step5: Correctly solve for x

So \( 6x + 3 + 5x - 1 = 90 \)
Combine like terms: \( 11x + 2 = 90 \)
Subtract 2: \( 11x = 88 \)
\( x = 8 \) (same x, but the angle sum for acute angles is 90). Then \( \angle A = 6(8) + 3 = 51^\circ \), \( \angle C = 5(8) - 1 = 39^\circ \). Now, triangle \( EFG \) is also a right triangle ( \( \angle F = 90^\circ \)) with a marked side equal to \( ABC \)'s marked side, so they are congruent by HL or AAS. Thus, \( \angle E = \angle A = 51^\circ \), so \( \angle G = 90 - 51 = 39^\circ \)? Wait, no—wait, in triangle \( EFG \), right angle at \( F \), so \( \angle E + \angle G = 90^\circ \). If \( \angle E = \angle A = 51^\circ \), then \( \angle G = 90 - 51 = 39^\circ \)? Wait, but \( \angle C \) is \( 39^\circ \), and since the triangles are congruent, \( \angle G = \angle C = 39^\circ \)? Wait, no, let's check again.

Wait, triangle \( ABC \): right angle at \( B \), legs with one marked (so isosceles? No, the marked side is one leg, so \( AB = BC \)? Wait, the mark is on \( AB \) and \( EF \), so \( AB = EF \). Triangle \( ABC \): right-angled at \( B \), so \( AB \) and \( BC \) are legs. Triangle \( EFG \): right-angled at \( F \), so \( EF \) and \( FG \) are legs. Since \( AB = EF \) (marked), and they are right triangles, maybe \( \angle A = \angle E \), so \( \angle C = \angle G \). Wait, \( \angle A = 6x + 3 \), \( \angle C = 5x - 1 \), and in right triangle, \( \angle A + \angle C = 90 \), so \( 6x + 3 + 5x - 1 = 90 \) → \( 11x + 2 = 90 \) → \( 11x = 88 \) → \( x = 8 \). Then \( \angle A = 6*8 + 3 = 51 \), \( \angle C = 5*8 - 1 = 39 \). Then in triangle \( EFG \), right-angled at \( F \), so \( \angle E + \angle G = 90 \). Since \( \angle E = \angle A = 51 \) (corresponding angles, congruent triangles), then \( \angle G = 90 - 51 = 39 \). Alternatively, since \( \angle C = 39 \), and \( \angle G = \angle C \) (congruent triangles), so \( \angle G = 39^\circ \).

Answer:

\( 39 \)