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If $A$ and $B$ are positive acute angles and $\cos A=\large\frac{1}{7},$$\cos B=\large\frac{13}{14}$ then find $A-B$

$(a)\;60^{\large\circ }\qquad(b)\;30^{\large \circ}\qquad(c)\;45^{\large\circ}\qquad(d)\;90^{\large\circ}$

1 Answer

Since $\cos A=\large\frac{1}{7}$
$\cos B=\large\frac{13}{14}$
$\sin A=\sqrt{1-\cos ^2A}$
$\qquad=\sqrt{1-\big(\large\frac{1}{7}\big)^2}$
$\qquad=\large\frac{4\sqrt 3}{7}$
$\sin A=\sqrt{1-\cos ^2B}$
$\qquad=\sqrt{1-\big(\large\frac{13}{14}\big)^2}$
$\qquad=\large\frac{3\sqrt 3}{14}$
Now,$\cos(A-B)=\cos A.\cos B+\sin A.\sin B$
$\Rightarrow \large\frac{1}{7}\frac{13}{14}+\frac{4\sqrt 3}{7}.\frac{3\sqrt 3}{14}$
$\Rightarrow \large\frac{13}{98}+\frac{12\times 3}{98}$
$\Rightarrow \large\frac{13+36}{98}$
$\Rightarrow \large\frac{49}{98}$
$\Rightarrow \large\frac{1}{2}$
$\cos(A-B)=\large\frac{1}{2}$
$A-B=\cos^{-1}\big(\large\frac{1}{2}\big)$
$\Rightarrow 60^{\large\circ}$
Hence (a) is the correct answer.
answered Oct 16, 2013 by sreemathi.v
 

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