Question

1. r (right triangle with 42°); 12) triangle with angles ((x+20)^circ), ((6x)^circ), ((3x)^circ); other triangles: (38^circ), (26) (handwritten), (39) (handwritten); sections: red, green, orange

Explanation:

Answer:

For problem 8 (right triangle with angle 42° and right angle, finding angle at R):
The sum of angles in a triangle is 180°. So angle at R is \( 180 - 90 - 42 = 48^\circ \).

For problem 12 (triangle with angles \( (x + 20)^\circ \), \( (6x)^\circ \), \( (3x)^\circ \)):
Sum of angles in a triangle is 180°, so \( (x + 20) + 6x + 3x = 180 \).
Combining like terms: \( 10x + 20 = 180 \).
Subtract 20: \( 10x = 160 \).
Divide by 10: \( x = 16 \).
Then angles: \( x + 20 = 36^\circ \), \( 6x = 96^\circ \), \( 3x = 48^\circ \).