In the figure, if m∠abd = 120º, then m∠adc =? º.

The figure that this question is referring to is attached. We must use the Law of Sines to solve this question, which is as follows:

a/sinA = b/sinB = c/sinC

This applies to any triangle. We are told that ∠ABD = 120º. We are asked to solve for ∠ADC. We know that ∠ADC + ∠ADB = 180º. If we assign ∠ADC = x, then ∠ADB = 180º - x. We can now apply the law of sines to this data.

35/sin120 = 30/sin(180-x)
sin(180-x) = (30/35)(sin120)
sin(180-x) = 0.742
sin-1(sin(180-x)) = sin-1(0.742)
180 - x = 48º
x = 132º

We have already assigned x = ∠ADC; therefore, ∠ADC = 132º.

Arhombus is a quadrilateral.