# In the triangle ADC (shown), AC is 194 cm. What is the length of BD (in triangle BDC) to the nearest centimetre? Question number 7....

**In the triangle ADC (shown), AC is 194 cm. What is the length of BD (in triangle BDC) to the nearest centimetre?**

**Question number 7.**

**http://staff.argyll.epsb.ca/jreed/math30p/correspondence/booklets/PureMath30AB_5A.pdf**

I know that tan15=y(DC)/194cm

**I know the angle BDC is 45 degrees (given); then angle B is also 45 degrees. I know we have a 45-45-90 triangle.**

**Also I know the lengths of the sides of 45-45-90 triangle are in the ratio of 1:1: `sqrt(2)` **

**Also I know the answer is one of the following: 72cm,73cm,74cm,75cm**

**Find the length of BD**

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### 1 Answer

To continue with your `tan 15` = `y/194` that is, `(DC)/194` in `Delta (ACD)`

`therefore y= tan 15 times 194`

`y= 51.98` ie DC = 51.98

You are correct to say that the ratio in the triangle BCD is `1 : 1: sqrt(2)`

Therefore if DC = 51.98 then BC = 51.98 (they have the same ratio)

As we already have a ratio of `1:1:sqrt(2)`

1 x 51.98 : 1 x 51.98 : `sqrt(2)` x 51.98

`therefore` BD = 73, 51 and rounded off that gives us 74 cm

`therefore`C (74 cm)