Question

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

1. value of θ.
2. length of its base AB, in nearest metre.

Answer 1

Given: In right triangle ABC, right-angled at B: AB = `sqrt(3)` × BC and AC inclined plane = 15 m.

Step-wise calculation:

1. Let BC = h. Then AB = `sqrt(3) xx h`.

2. Use Pythagoras on triangle ABC:

AC² = AB² + BC²

`15^2 = (sqrt(3) xx h)^2 + h^2`

= 3h² + h²

= 4h²

3. Solve for h:

225 = 4h²

⇒ h² = 56.25

⇒ h = 7.5 m

4. Find AB:

`AB = sqrt(3) xx h` 

= `sqrt(3) xx 7.5` 

= 1.732 × 7.5

= 12.99 m

= 13 m   ...(Nearest metre)

5. Find θ angle at A:

`sin θ = "Opposite"/"Hypotenuse"`

= `(BC)/(AC)`

= `7.5/15`

= `1/2`

⇒ θ = 30°

Equivalently tan θ = `(BC)/(AB)`

= `1/sqrt(3)`

⇒ θ = 30°