Question

In the figure, if m∠ABD = 120º, then m∠ADC = ___

Answer 1

Okay first, use the sin rule ($\frac{SinA}{a} = \frac{SinB}{b}$
so $\frac{Sin120}{35} * 30 = SinB$
sin^-1 this and you get 47.928....this is angle ADB
and as it is on a straight line, 180 - this angle gets ADC
180 - 47.928.... = 132.07

Answer 2

Answer-

$\boxed{\boxed{m\angle ADC=132^{\circ}}}$

Solution-

Law of Sines-

$\dfrac{a}{\sin A} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$

Here,

B = 120°

b = 35 units

d = 30 units

Let us assume that m∠ADC be x, so m∠ADB=180-x (as they are complementary angles)

Applying Laws of sine for ΔABD,

$\Rightarrow \dfrac{b}{\sin B}=\dfrac{d}{\sin D}$

$\Rightarrow \dfrac{35}{\sin 120}=\dfrac{30}{\sin (180-x)}$

$\Rightarrow \sin (180-x)=\dfrac{30\times \sin 120}{35}$

$\Rightarrow \sin (180-x)=\dfrac{3\sqrt{3}}{7}=0.742$

$\Rightarrow 180-x=\sin^{-1}0.742$

$\Rightarrow 180-x=47.9\approx 48$

$\Rightarrow x=180-48=132^{\circ}$