Question

In the following figure, CD = 20 cm, find AB and BC.

Answer 1

Given:

CD = 20 cm.

AB ⟂ BD at B, so AB is vertical and BD is horizontal.

C and D lie on BD with C between B and D.

∠BCA = 60° and ∠CDA = 45° angles at C and D formed by the rays to A.

Step-wise calculation:

1. Put coordinates: B(0, 0). 

Let BC = c, so C(c, 0) and D(c + 20, 0). 

Let A be (0, h) so AB = h.

2. Angle at C: The angle between the leftward horizontal CB and CA equals 60°. 

So, `tan 60^circ = h/c`

⇒ `sqrt(3) = h/c`

⇒ `h = csqrt(3)`   ...(1)

3. Angle at D: The angle between the leftward horizontal DC and DA equals 45°. 

So, `tan 45^circ = h/(c + 20)`

⇒ `1 = h/(c + 20)`

⇒ h = c + 20   ...(2)

4. Equate (1) and (2):

`csqrt(3) = c + 20`

`c(sqrt(3) - 1) = 20` 

`c = 20/(sqrt(3) - 1)` 

= `(20(sqrt(3) + 1))/(3 - 1)` 

= `10(sqrt(3) + 1) . (BC)`

5. Then h = c + 20

= `10(sqrt(3) + 1) + 20` 

= `10sqrt(3) + 30` 

= `10(3 + sqrt(3)) . (AB)`

BC = `10(sqrt(3) + 1)` cm

= 27.3205 cm

AB = `10(3 + sqrt(3))` cm 

= `30 + 10sqrt(3)` cm 

= 47.3205 cm

CD = 20 cm, as given.