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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 = 180<span>º - 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 = 48</span><span>º
x = 132</span><span>º
</span>
We have already assigned x = ∠ADC; therefore, ∠ADC = 132<span>º.</span>
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 = 180<span>º - 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 = 48</span><span>º
x = 132</span><span>º
</span>
We have already assigned x = ∠ADC; therefore, ∠ADC = 132<span>º.</span>
