An inclined plane AC is prepared with its base AB which is √3 times its vertical height BC. The length of the inclined plane is 15 m. Find:

(a) value of θ.

(b) length of its base AB, in nearest metre.

(a) According to question,

⇒ AB = $\sqrt{3} \times BC$

⇒ AB = $\sqrt{3}BC$

⇒ BC = $\dfrac{AB}{\sqrt{3}}$

From figure,

⇒ tan θ = $\dfrac{BC}{AB}$

⇒ tan θ = $\dfrac{BC}{\sqrt{3}BC}$

⇒ tan θ = $\dfrac{1}{\sqrt{3}}$

⇒ tan θ = tan 30°

⇒ θ = 30°.

Hence, θ = 30°.

(b) In right angle triangle ABC,

By pythagoras theorem,

⇒ AC² = BC² + AB²

⇒ 15² = $\Big(\dfrac{AB}{\sqrt{3}}\Big)^2$ + AB²

⇒ 225 = $\dfrac{AB^2}{3}$ + AB²

⇒ 225 = $\dfrac{AB^2 + 3AB^2}{3}$

⇒ 225 = $\dfrac{4AB^2}{3}$

⇒ AB² = $\dfrac{225 \times 3}{4}$

⇒ AB² = $\dfrac{675}{4}$

⇒ AB² = 168.75

⇒ AB = $\sqrt{168.75}$

⇒ AB = 12.99 m

Rounding off,

AB = 13 m.

Hence, AB = 13 m.