In △ABC, the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1. Then cos(A3) is equal to

We are given that the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1⇒∠BAD∠CAD=21⇒∠BAD = 2∠CAD...................… (1)Since we know ∠BAC=∠BAD+∠CADSubstitute the values in equation (1)⇒∠BAC = 2∠CAD+∠CAD⇒∠BAC = 3∠CADWe write ∠BAC=A⇒A = 3∠CADDivide both sides by 3⇒A3=∠CAD..................… (2)Substitute value from equation (2) in (1) ⇒∠BAD=2A3..................… (3)Now we apply the law of sine in both triangles made by the median separately.In △BADBDsin∠BAD=ADsinB⇒a2sin2A3=ADsinB ⇒a2sinBsin2A3=AD .................… (4)In △CADCDsin∠CAD=ADsinC ⇒a2sinA3=ADsinC ⇒a2sinCsinA3=AD ....................… (5)Since AD is the median and will have fixed length, we equate values from (4) and (5)⇒a2sinBsin2A3=a2sinCsinA3 ⇒sinBsin2A3=sinCsinA3 Since, sin2x=2sinxcosx ⇒sinB2sinA3cosA3=sinCsinA3 ⇒sinB2cosA3=sinC1 ⇒sinB2sinC=cosA3 ∴ The value of cosA3 is equal to sinB2sinC∴ Option 1 is correct.

In △ABC, the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1. Then cos(A3) is equal to

We are given that the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1⇒∠BAD∠CAD=21⇒∠BAD = 2∠CAD...................… (1)Since we know ∠BAC=∠BAD+∠CADSubstitute the values in equation (1)⇒∠BAC = 2∠CAD+∠CAD⇒∠BAC = 3∠CADWe write ∠BAC=A⇒A = 3∠CADDivide both sides by 3⇒A3=∠CAD..................… (2)Substitute value from equation (2) in (1) ⇒∠BAD=2A3..................… (3)Now we apply the law of sine in both triangles made by the median separately.In △BADBDsin∠BAD=ADsinB⇒a2sin2A3=ADsinB ⇒a2sinBsin2A3=AD .................… (4)In △CADCDsin∠CAD=ADsinC ⇒a2sinA3=ADsinC ⇒a2sinCsinA3=AD ....................… (5)Since AD is the median and will have fixed length, we equate values from (4) and (5)⇒a2sinBsin2A3=a2sinCsinA3 ⇒sinBsin2A3=sinCsinA3 Since, sin2x=2sinxcosx ⇒sinB2sinA3cosA3=sinCsinA3 ⇒sinB2cosA3=sinC1 ⇒sinB2sinC=cosA3 ∴ The value of cosA3 is equal to sinB2sinC∴ Option 1 is correct.

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In △ABC, the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1. Then cos( $\frac{A}{3}$ ) is equal to

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\frac{sinB}{2 sinC}

\frac{sinC}{2 sinB}

\frac{2 sinB}{sinC}

None of These

answer is A.

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Detailed Solution

We are given that the median AD divides ∠BAC such that ∠BAD:∠CAD=2:1

\Rightarrow \frac{∠ BAD}{∠ CAD} = \frac{2}{1}

⇒∠BAD = 2∠CAD...................… (1)
Since we know ∠BAC=∠BAD+∠CAD
Substitute the values in equation (1)
⇒∠BAC = 2∠CAD+∠CAD
⇒∠BAC = 3∠CAD
We write ∠BAC=A
⇒A = 3∠CAD
Divide both sides by 3