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.
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)