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In a ∆ ABC, ∠B = (π / 3) , ∠C = (π / 4) and D divides BC internally in the ratio 1 : 3 then sin (sin ∠BAD / sin∠CAD) is equal to
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1 √2 -
1 √3 -
1 √6 - √6
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Correct Option: C
| ∠B = | , ∠ = | 3 | 4 |
| and | = | DC | 3 |
From ∆ ABD,
| = | sin BAD | sin ABD |
| ⇒ | = | sin BAD | sin (π / 3) |
| ⇒ | = | sin BAD | (√3 / 2) |
| ⇒ AD = | . | ...(i) | 2 | sin BAD |
From ∆ ADC,
| = | sin DAC | sin ACD |
| ⇒ | = | sin DAC | sin (π / 4) |
| ⇒ AD = | . | ...(ii) | √2 | sin DAC |
From equations (i) and (ii),
| . | = | . | 2 | sin BAD | √2 | sin DAC |
| = | - | × √2 × | sin DAC | 2 | 3 |
| = | = | √2 × √3 | √6 |
