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

1. A
$\frac{\mathit{sinB}}{2\mathit{sinC}}$
2. B
$\frac{\mathit{sinC}}{2\mathit{sinB}}$
3. C
$\frac{2\mathit{sinB}}{\mathit{sinC}}$
4. D
None of These

Fill Out the Form for Expert Academic Guidance!l

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

Solution:

$⇒\frac{\angle \mathit{BAD}}{\angle \mathit{CAD}}=\frac{2}{1}$
Substitute the values in equation (1)
We write  ∠BAC=A
Divide both sides by 3

Substitute value from equation (2) in (1)

Now we apply the law of sine in both triangles made by the median separately.
$\frac{\mathit{BD}}{\mathit{sin}\left(\angle \mathit{BAD}\right)}=\frac{\mathit{AD}}{\mathit{sinB}}$

Since AD is the median and will have fixed length, we equate values from (4) and (5)

Since, sin2x=2sinxcosx

∴ Option 1 is correct.

Related content

 Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formula Perimeter of Triangle Formula Area Formulae Volume of Cone Formula Matrices and Determinants_mathematics Critical Points Solved Examples Type of relations_mathematics

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)