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m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

is there a question here?

step-by-step explanation:

b) k’(6, 4); l’(3, 3)

step-by-step explanation:

a 270° clockwise rotation maps every point (x, y) to (y, -x). (it negates the x-coordinate and then switches it and the y-coordinate.)

this means the endpoints of the side congruent to kl, k' and l', would be mapped as:

k'(-4, 6)→(6, 4)

l'(-3, 3)→(3, 3)