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Deffense [45]
4 years ago
6

70 POINTS!!!!!!!!!! Given: E, F, Q, D∈k(O),O ∈ ED, m∠DFQ = 10°, measure of arc EF = 28° Find: Angles of △EFQ

Mathematics
1 answer:
Likurg_2 [28]4 years ago
5 0

Answer:

The measures of angles of triangle EFQ are

1) m\angle EFQ=100\°

2) m\angle FEQ=66\°

3) m\angle EQF=14\°

Step-by-step explanation:

step 1

Find the measure of arc QD

we know that

The inscribed angle is half that of the arc it comprises.

m\angle DFQ=\frac{1}{2}(arc\ QD)

substitute the given value

10\°=\frac{1}{2}(arc\ QD)

20\°=(arc\ QD)

arc\ QD=20\°

step 2

Find the measure of arc FQ

we know that

arc\ QD+arc\ FQ+arc\ EF=180\° ---> because ED is a diameter (the diameter divide the circle into two equal parts)

substitute the given values

20\°+arc\ FQ+28\°=180\°

arc\ FQ=180\°-48\°=132\°

step 3

Find the measure of angle EFQ

we know that

The inscribed angle is half that of the arc it comprises.

m\angle EFQ=\frac{1}{2}(arc\ QD+arc\ ED)

substitute the given value

m\angle EFQ=\frac{1}{2}(20\°+180\°)=100\°

step 4

Find the measure of angle FEQ

we know that

The inscribed angle is half that of the arc it comprises.

m\angle FEQ=\frac{1}{2}(arc\ FQ)

substitute the given value

m\angle FEQ=\frac{1}{2}(132\°)=66\°

step 5

Find the measure of angle EQF

we know that

The inscribed angle is half that of the arc it comprises.

m\angle EQF=\frac{1}{2}(arc\ EF)

substitute the given value

m\angle EQF=\frac{1}{2}(28\°)=14\°

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This will be easier to write, and a lot easier to read, if we temporarily
use another symbol ... say, 'Q' ... to represent  ' sin(2x) ' .
Here we go:

Original equation:                Q² - 0.5 Q  =  0

Factor the left side:            Q (Q - 0.5)  =  0

This equation is true if either factor is zero:


--  If  Q=0, then  sin(2x) = 0

                                 2x = 0°,  180°,  360°

                                   x = 0°,  90°,  180°


-- If  (Q-0.5) = 0, then      Q  =  0.5

                                sin(2x)  =  0.5          

                                       2x  =  30°,  150°

                                         x  =  15°,  75°


The whole collection of solutions
between  0°  and  360° :

          x = 0°,  15°,  75°,  90°,  180° .

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