In the given figure, ∠B = 60°, AB = 16 cm and BC = 23 cm. Calculate :

(i) BE

(ii) AC.

(i) In Δ ABE,

$\text{cos 60°} = \dfrac{Base}{Hypotenuse}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{BE}{AB}\\[1em] ⇒ \dfrac{1}{2} = \dfrac{BE}{16}\\[1em] ⇒ BE = \dfrac{16}{2} = 8 cm$

Hence, BE = 8 cm.

(ii) In Δ ABE, according to Pythagoras theorem,

⇒ AB² = BE² + EA² (∵ AB is hypotenuse)

⇒ 16² = 8² + EA²

⇒ 256 = 64 + EA²

⇒ EA² = 256 - 64

⇒ EA² = 192

⇒ EA = $\sqrt{192}$

⇒ EA = 8 $\sqrt{3}$

EC = BC - BE

= 23 - 8 cm

= 15 cm

In Δ AEC, according to Pythagoras theorem,

⇒ AC² = AE² + EC² (∵ AC is hypotenuse)

⇒ AC² = (8 $\sqrt{3}$)² + 15²

⇒ AC² = 192 + 225

⇒ AC² = 417

⇒ AC = $\sqrt{417}$ = 20.42 cm

Hence, AC = 20.42 cm.