Geometry Problem 1505. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.
Details: Click on the figure below.
More Details
To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.
To post a solution to this problem click Comment underneath the post, or click into the line that says, “Enter Comment.” Type what you want to say and press Publish to post your solution.
Draw line OD //L1
ReplyDeletewe have ^(COD)= 30 and ^(AOD)= 100 so ^(AOB)=^(BOC)= 35
so ^(OBL2)= 180- 65= 115
Let L3C meet OB at C1 and let L1A meet OB at A1
ReplyDeleteLet y = <COC1 = <AOA1
Now <OC1C = < B
Hence y + 80 = B....(1) and y + 30 + B = 180....(2)
(1) -(2) ; 80 - B - 30 = B - 180
Therefore x = 115
Sumith Peiris
Moratuwa
Sri Lanka