In the figures given below, find AB :

(i)

(ii)

(iii)

(i)

In Δ BCD,

tan 45° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

⇒ 1 = $\dfrac{DC}{BC}$

⇒ 1 = $\dfrac{80}{BC}$

⇒ BC = 80 m

In Δ ADC,

tan 30° = $\dfrac{\text{Perpendicular}}{\text{Base}}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{DC}{AC}$

⇒ $\dfrac{1}{\sqrt{3}} = \dfrac{80}{AC}$

⇒ AC = 80 $\sqrt{3}$ m

Now, AB = AC - BC

= 80 $\sqrt{3}$ - 80 m

= 80($\sqrt{3}$ - 1) m

= 80(1.732 - 1) m

= 80 x 0.73 m

= 58.56 m

Hence, the value of AB = 58.56 m.

(ii)

Given: CD = AC = 40 m

In Δ ABC,

sin 60° = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{AB}{AC}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{AB}{40}$

⇒ AB = 40 $\dfrac{\sqrt{3}}{2}$ m = 20 $\sqrt{3}$ m = 34.64 m

Hence, the value of AB = 34.64 m.

(iii) Given: AB = BD

⇒ ∠ ADB = ∠ DAB (angles corresponding to the equals sides are always equal)

In triangle ABD, sum of all angles of triangle is 180°.

⇒ ∠ ADB + ∠ DAB + ∠ ABD = 180°

⇒ 30° + 30° + ∠ ABD = 180°

⇒ 60° + ∠ ABD = 180°

⇒ ∠ ABD = 180° - 60° = 120°

And, ∠ ABD = ∠ BDC + ∠ BCD (exterior angle property)

⇒ 120° = ∠ BDC + 90°

⇒ ∠ BDC = 120° - 90° = 30°

In Δ DBC,

cos 30° = $\dfrac{\text{Base}}{\text{Hypotenuse}}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{DC}{BD}$

⇒ $\dfrac{\sqrt{3}}{2} = \dfrac{20}{BD}$

⇒ BD = $\dfrac{20 \times 2}{\sqrt{3}}$

⇒ BD = $\dfrac{40}{\sqrt{3}}$ = 23.1m

From the figure, BD = AB = 23.1 m

Hence, the value of AB = 23.1 m.