Question

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?

Answer 1

The measure of angle AED is
`94(8)/(11)\°`

Explanation:

step 1

Find the measure of x

we know that

The measure of the interior angle is the semi-sum of the arcs comprising it and its opposite

In this problem

m<AEC is a interior angle

so

`m<AEC=(1)/(2)(arc\ AC+arc\ DB)`

substitute the values and solve for x

`7x\°=(1)/(2)(134\°+3x\°)`

`14x\°=(134\°+3x\°)`

`14x\°-3x\°=134\°`

`11x\°=134\°`

`x=(134/11)\°`

step 2

Find the measure of angle AED

we know that

`m<AEC+m<AED=180\°` -----> by supplementary angles

`m<AED=180\°-m<AEC`

`m<AED=180\°-7x`

`m<AED=180\°-7(134/11)\°`

`m<AED=180\°-(938/11)\°`

`m<AED=(1,042/11)\°`

Convert to mixed number

`(1,042/11)\°=(1,034/11)\°+(8/11)\°=94\°+(8/11)\°=94(8)/(11)\°`