Question

In a triangle $ABC$, $P$ is an interior point such that $\angle PAB=10$, $\angle PBA=20$, $\angle PCA=30$ And $\angle PAC=40$. $a$,$b$,$c$ are the sides of triangle $ABC$.
Value of $\cos (A-B)+\cos (B-C)+\cos (C-A)$ is equal to

• A. $\frac{3}{2}$
• B. $\frac{\sqrt{3}}{2}+1$
• C. $1+\sqrt{3}$
• D. $\sqrt{3}+\frac{1}{2}$

Answer 1

The correct option is C $1+\sqrt{3}$

From the picture we have using $\sin$ law

$\frac{AP}{\sin {20}^{\circ }}.\frac{BP}{\sin ({80}^{\circ }-x)}.\frac{CP}{\sin {40}^{\circ }}$$=\frac{AP}{\sin {30}^{\circ }}.\frac{BP}{\sin {10}^{\circ }}.\frac{CP}{\sin x}$

or, $\sin {20}^{\circ }.\sin ({80}^{\circ }-x).\sin {40}^{\circ }=\sin {30}^{\circ }.\sin {10}^{\circ }\sin x$

or, $2\sin {10}^{\circ }.\cos {10}^{\circ }$$.\sin ({80}^{\circ }-x).\sin {40}^{\circ }=\frac{1}{2}.$$\sin {10}^{\circ }\sin x$

or, $4\cos {10}^{\circ }$$.\sin ({80}^{\circ }-x).\sin {40}^{\circ }=\sin x$

$x={60}^{\circ }$ satisfies both sides.

As $4\cos {10}^{\circ }$$.\sin ({80}^{\circ }-{60}^{\circ }).\sin {40}^{\circ }$

$=4\cos {10}^{\circ }$$.\sin {20}^{\circ }.\sin {40}^{\circ }$

$=(2\sin {20}^{\circ }.\cos {10}^{\circ }).\left(2\sin {40}^{\circ }\right)$

$=[\sin {30}^{\circ }+\sin {10}^{\circ }].\left(2\sin {40}^{\circ }\right)$

$=\sin {40}^{\circ }+2\sin {10}^{\circ }.\sin {40}^{\circ }$

$=\sin {40}^{\circ }+\cos {30}^{\circ }-\cos {50}^{\circ }$

$=\sin {40}^{\circ }+\cos {30}^{\circ }-\sin {40}^{\circ }$

$=\cos {30}^{\circ }=\sin {60}^{\circ }$.

$\therefore \angle$ PBC $={60}^{\circ }$ and $\angle$PCB$={30}^{\circ }$.

$\angle$A $={50}^{\circ }$, $\angle$ B$={80}^{\circ }$ and $\angle$C$={50}^{\circ }$.

$\therefore \cos (A-B)+\cos (B-C)+\cos (C-A)=\cos {30}^{\circ }+\cos {30}^{\circ }+\cos 0^{\circ }$ [Since $\cos (-{30}^{\circ })=\cos {30}^{\circ }$]

$=2\cos {30}^{\circ }+1$

$=\sqrt{3}+1$.