1.

If `P`is a point on the altitude AD of the triangle ABC such the `/_C B P=B/3,`then AP is equal to`2asinC/3`(b) `2bsinC/3``2csinB/3`(d) `2csinC/3`

Answer» We can create a diagram using the given details.
Please refer to video to see the diagram.
From the diagram,
`/_ABP = (2B)/3 and /_APB = 90+B/3`
In `Delta APB`,
`(AP)/(sin((2B)/3)) = c/(sin(90+B/3))`
`=>(AP)/(2sin(B/3)cos(B/3)) = c/(cos(B/3))`
`=>AP = 2csin(B/3)`


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