We appreciate your visit to In the diagram PQRS JQK and LRK are straight lines Angle angle JQK 2y circ Angle angle QKL x circ Angle angle KLM 33 circ. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Answer:
∠JKL = 38°
Step-by-step explanation:
I did the question in class today!
Thanks for taking the time to read In the diagram PQRS JQK and LRK are straight lines Angle angle JQK 2y circ Angle angle QKL x circ Angle angle KLM 33 circ. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
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°