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.
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
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)