Question

23 questions please help me answer

Answer 1

GIVEN=>

in the photo lines AD, BE and CF pas through the point of.

/_BOC=(4x°) --(1)

/_COD= y°

/_BOA= 30°

/_ FOA=(2x°) --(2)

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find value=>

(a) /_BOC=?

(b) /_COD=?

(c) /_DOE=?

(d) /_EOF=?

(e)/_ FOA=?

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SOLUTION=>

LINE FC ONE SIND ON ANGLE VALUE 180°(linear pair angle)

=/_BOC+BOA+/_FOA=180°(linear pair angle)

$= > 4x + 30 + 2x = 180 \\ \\ = > 6x = 180 - 30 \\ \\ = > x = \frac{150}{6} \\ \\ = > x = 25$

now x value eq(1) put

/_BOC=(4x°)

/_BOC=(4×25)

/_BOC=100°

now x value eq(2) put

/_ FOA=(2x°)

/_ FOA=(2×25)

/_ FOA=50°

/_COD( y°) =/_ FOA=(2x°)=50° (heading angle)

/_DOE=/_BOA= 30°(heading angle)

/_EOF=/_BOC(4x°)=100°(heading angle)

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answer=>

(a) /_BOC=100°

(b) /_COD=50°

(c) /_DOE=30°

(d) /_EOF=100°

(e)/_ FOA=50°

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