In the diagram shown, chords AB and CD intersect at E. The measure of (AC) ̂ is 134°, the measure of (DB) ̂ is (3x)° and the measure of ∠AEC is (7x)°. What is the degree measure of ∠ AED?
Accepted Solution
A:
Answer:The measure of angle AED is [tex]94\frac{8}{11}\°[/tex] Step-by-step explanation:step 1Find the measure of xwe know thatThe measure of the interior angle is the semi-sum of the arcs comprising it and its oppositeIn this problemm<AEC is a interior angleso[tex]m<AEC=\frac{1}{2}(arc\ AC+arc\ DB)[/tex]substitute the values and solve for x[tex]7x\°=\frac{1}{2}(134\°+3x\°)[/tex][tex]14x\°=(134\°+3x\°)[/tex][tex]14x\°-3x\°=134\°[/tex][tex]11x\°=134\°[/tex][tex]x=(134/11)\°[/tex]step 2 Find the measure of angle AEDwe know that[tex]m<AEC+m<AED=180\°[/tex] -----> by supplementary angles[tex]m<AED=180\°-m<AEC[/tex][tex]m<AED=180\°-7x[/tex][tex]m<AED=180\°-7(134/11)\°[/tex] [tex]m<AED=180\°-(938/11)\°[/tex] [tex]m<AED=(1,042/11)\°[/tex] Convert to mixed number[tex](1,042/11)\°=(1,034/11)\°+(8/11)\°=94\°+(8/11)\°=94\frac{8}{11}\°[/tex]