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ABCD is a rectangle where the ratio of the length of AB and BC is 3 : 2. If P is the mid-point of AB, then the value of sin ∠CPB is
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3 5 -
2 5 -
3 4 -
4 5
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Correct Option: D
As per the given in question , we draw a figure rectangle ABCD
Given , AB : BC = 3 : 2
Let AB = 3y units and BC = 2y units
| ⇒ PB = | y units. | |
| 2 |
From ∆ PCB ,
CP = √PB² + BC²
| CP = √ | + 4y² | |
| 4 |
| CP = √ | = | units | ||
| 4 | 2 |
| ∴ sin ∠CPB = | |
| CP |
| sin ∠CPB = | |
| 2 |
| sin ∠CPB = | |
| 5 |
Take AB = 6y and BC = 4y ⇒ BP = 3y ⇒ CP = 5y
| and sin ∠BPC = | = | = | |||
| CP | 5y | 5 |
