Question

In the adjoining figure, AB = 30 cm, find AD and BC.

Answer 1

Given:

AB = 30 cm.

A, B, C are collinear, AC is the base, AD ⟂ AB, so D is directly above A.

∠(BD, BA) = 45°, so BD makes 45° above the left horizontal and ∠(DC, CB) = 30°, so DC makes 30° above the left horizontal.

Step-wise calculation:

1. Put A at (0, 0), so B = (30, 0). 

Let BC = b, so C = (30 + b, 0).

Since AD is vertical above A, D = (0, h) where h = AD.

2. Line BD makes 45° above the left horizontal its slope = tan (135°) = –1. 

Equation through B(30, 0):

y = –(x – 30) 

= –x + 30

3. Line CD makes 30° above the left horizontal its slope = `tan 150^circ = -1/sqrt(3)`. 

Equation through C(30 + b, 0):

`y = -(1/sqrt(3)) xx (x - (30 + b))`

4. D is the intersection of BD and CD and lies above A, so x_D = 0. 

Plug x = 0 into the BD equation to get: y_D = –0 + 30 = 30. 

Hence, D = (0, 30) and AD = h = 30 cm.

5. Use D(0, 30) on the CD line to find b:

Slope = `(30 - 0)/(0 - (30 + b))`

= `-30/(30 + b)` must equal `-1/sqrt(3)`

So, `30/(30 + b) = 1/sqrt(3)`

⇒ `30 + b = 30sqrt(3)` 

⇒ `b = 30(sqrt(3) - 1)`

6. Therefore, BC = `30(sqrt(3) - 1)` cm

= 21.96 cm

AD = 30 cm.

BC = `30(sqrt(3) - 1)` cm (≈ 21.96 cm).