This is apart of my semester exam if you don’t know it please don’t answer it
1 answer:
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Answer:
∠JKL = 38°
Step-by-step explanation:
PQRS, JQK and LRK are straight lines
Let's take the straight lines in the diagrams one after the other to find what they consist.
The related diagram can be found at brainly (question ID: 18713345)
Find attached the diagram used for solving the question.
For straight line PQRS,
2x°+y°+x°+2y° = 180°
(Sum of angles on a Straight line = 180°)
Collect like terms
3x° + 3y° = 180°
Also straight line PQRS = straight line PQR + straight line SRQ
For straight line PQR,
2y + x + ∠RQM = 180° ....equation 1
For straight line SRQ,
2x + y + ∠MRQ = 180° ....equation 2
Straight line PQRS = addition of equation 1 and 2
By collecting like times
3x +3y + ∠RQM + ∠MRQ = 360°....equation 3
Given ∠QMR = 33°
∠RQM + ∠MRQ + ∠QMR = 180° (sum of angles in a triangle)
∠RQM + ∠MRQ + 33° = 180°
∠RQM + ∠MRQ = 180-33
∠RQM + ∠MRQ = 147° ...equation 4
Insert equation 4 in 3
3x° +3y° + 147° = 360°
3x +3y = 360 - 147
3x +3y = 213
3(x+y) = 3(71)
x+y = 71°
∠JQP = ∠RQK = 2y° (vertical angles are equal)
∠LRS = ∠QRK = 2x° (vertical angles are equal)
∠QRK + ∠RQK + ∠QKR = 180° (sum of angles in a triangle)
2x+2y + ∠QKR = 180
2(x+y) + ∠QKR = 180
2(71) + ∠QKR = 180
142 + ∠QKR = 180
∠QKR = 180 - 142
∠QKR = 38°
∠JKL = ∠QKR = 38°
