Question

In fig if CD = 2BC then value of BAD is ​

Answer 1

∠BAD = 60°

Step-by-step explanation:

• Given data

        ∠BAC = 30°

        CD = 2 BC      ...1)

• We can write

         BD = BC + CD

        BD = BC + 2 BC

        So

       BD = 3 BC       ...2)

• From triangle ΔABC

        $\mathbf{\tan 30^{\circ}=\frac{BC}{AB}}$        ...3)

• From triangle ΔABD

        $\mathbf{\tan \angle BAD=\frac{BD}{AB}}$        

• We can also write BD in terms of BC

        $\mathbf{\tan \angle BAD=\frac{3BC}{AB}}$         ...4)

• Divide equation 4) by equation 3)

        $\mathbf{\frac{\tan \angle BAD}{\tan 30^{\circ}}=\frac{\frac{3BC}{AB}}{\frac{BC}{AB}}}$

       So

        $\mathbf{\frac{\tan \angle BAD}{\tan 30^{\circ}}=3}$

• $\mathbf{\tan \angle BAD=3\tan 30^{\circ}=3\times \frac{1}{\sqrt{3}}=\sqrt{3}}$

        So

         $\mathbf{\tan \angle BAD=\tan 60^{\circ}}$

         ∠BAD = 60°

Answer 2

∠ BAD = 60°

Step-by-step explanation:

See the attached diagram.

Given that CD = 2BC, so if BC = h, then CD = 2h

From the right triangle Δ ABC,

$\tan 30^{\circ} = \frac{BC}{AB} = \frac{h}{x}$

⇒ h = x tan 30° ............ (1)

Now, from the right triangle Δ ABD,

$\tan BAD = \frac{BD}{AB} = \frac{3h}{x} = \frac{3 \times x\tan 30^{\circ}}{x} = \sqrt{3}$ {From equation (1)}

⇒ ∠ BAD = 60° (Answer)