Find all the angles formed at the intersection of two parallel secant lines if one of them is 42 degrees.
Let AB and CD be parallel lines, line MN – secant, K – point of intersection of line AB with MN, T – point of intersection of line CD with MN, angle AKM = 42 degrees.
When two parallel secant lines intersect, 8 angles are formed.
1. Angles AKM and BKM are adjacent. The sum of adjacent catches is 180 degrees, then:
angle AKM + angle BKM = 180 degrees;
42 + angle BKM = 180;
angle BKM = 180 – 42;
angle BKM = 138 degrees.
2. The corners AKM and NKB are vertical. The vertical angles are equal, then:
angle AKM = angle NKB;
angle NKB = 42 degrees.
3. The corners BKM and AKN are vertical. The vertical angles are equal, then:
angle BKM = angle AKN;
angle AKN = 138 degrees.
4. Angles AKN (AKT) and KTD – lying crosswise. The angles lying crosswise are equal, then:
angle AKT = angle KTD;
angle KTD = 138 degrees.
5. The corners KTD and CTN are vertical. The vertical angles are equal, then:
angle KTD = angle CTN;
angle CTN = 138 degrees.
6. Angles AKM and СTK – respectively. The corresponding angles are equal, then:
angle AKM = angle СTK;
angle СTK = 42 degrees.
7. The corners СTK and NTD are vertical. The vertical angles are equal, then:
angle СTK = angle NTD;
angle NTD = 42 degrees.
Answer: BKM angle = 138 degrees, NKB angle = 42 degrees, AKN angle = 138 degrees, KTD angle = 138 degrees, CTN angle = 138 degrees, СTK angle = 42 degrees, NTD angle = 42 degrees.
