− 1/2 (−3y+10) how do i work it out?
1 answer:
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The value of x is 120°
Explanation:
It is given that ABC is a straight line.
Also, ∠CBE = 20° and ∠CAF = 40°
We need to determine the value of ∠DBE
Let ∠DBE = x
Since, ∠CAF and ∠ABD are alternate interior angles.
By the alternate interior angles theorem, "if two parallel lines are cut by a transversal, then the pairs of alternate interior angles are equal".
Thus, ∠CAF and ∠ABD are equal.
Hence, ∠CAF = ∠ABD = 40°
Since, we know that, the angles in a straight line add up to 180°, we have,
∠ABD + ∠DBE + ∠CBE = 180°
Substituting the values, we get,
40° + x + 20° = 180°
Adding, we have,
60° + x = 180°
Subtracting both sides by 60, we have,
x = 120°
Thus, the value of x is 120°
