Answer:
Explanation:
The given diagram shows two points labeled B. Rename the one on segment CD as E.
We assume that AE is supposed to be the bisector of angle A. Since triangle ABC is isosceles, segment AE will be perpendicular to BC. In that case, triangle EBC is also isosceles, which means BE is also 6 cm.
Angle ABC is exterior to triangle BCD, so has a measure that is the sum of angles BDC and BCD. That is, angle ABC is 2y+y=3y. Angle CBE is also y, so angle ABE will be 3y+y=4y. This exterior angle to triangle BED is the sum of angles BED and BDE. We know BDE is 2y, so BED is 4y-2y=2y, which means triangle BED is isosceles and BD=BE=EC=x=6 cm.
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Finding the measure of angle y is a bit tricky. We know that angle ABC is 3y, and angle EBC is y. If we say X is the midpoint of BC, where AE crosses BC, then we can write the trig relations ...
BX=BA·cos(3y)
BX=BE·cos(y)
Filling in the measures of BA and BE, we can equate expressions for BX to get ...
8cos(3y) = 6cos(y)
Using the trig identity for cos(3y), we can write this as ...
8(4cos(y)³ -3cos(y)) = 6cos(y)
32cos(y)³ -30cos(y) = 0
2cos(y)(16cos(y)² -15) = 0
This has solutions cos(y)=0 and cos(y) = ±√(15/16). We are interested in the positive solution only:
y = arccos(√15/16) ≈ 14.478°
The value of y is about 14.5°.