Question

Given right triangle ABC with legs BC = 3, AC = 4. What is the length of the shorter angle trisector from C to the hypotenuse ?

Answer 1

Given `DeltaABC` a right triangle

where `angle ACB=90^@,BC=3cm and AC=4cm`

CD and CE are two trisectors of `angle ACB`,

CD which is closure to smaller leg will be the smaller trisector,

Now we have

`AB=sqrt(BC^2+AC^2)=sqrt(3^2+4^2)=5`

Now `sinA=(BC)/(AB)=3/5`

And
`angle ADC=180^@-angleACD-A=180^@-60^@-A`

So `sin angle ADC`

`=sin(120^2-A)`

`=sin120^@cosA-cos120^@sinA`

`=sqrt3/2xx4/5-(-1/2)xx3/5=(4sqrt3+3)/10`

Now by sine law for `Delta ACD` we have

`(CD)/sinA=(AC)/(sin angle ADC)`

`=>CD=(AC)/(sin angle ADC)xxsinA`

`=>CD=4/((4sqrt3+3)/10)xx3/5=24/(4sqrt3+3)~~2.4` cm