Question

solve only b. and don't say that answer is North west. I know that. .... but please find. exactly. using formula. ​

Answer 1

We're asked to find in which direction the city C is situated relative to C.

For this we've to find the angle made by the vector AC with vector AB. This angle means by what inclination the city C is situated along north of west.

We know, if R is the resultant of two vectors P and Q, then the angle made by R with P is,

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{Q\sin\theta}{P+Q\cos\theta}\right)}$

where $\theta$ is the angle between P and Q.

In this question,

• $\sf{P=AB=200\ km}$

• $\sf{Q=BC=300\ km}$

• $\sf{\theta=30^o}$

Then angle made by AC with AB is,

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{BC\sin\theta}{AB+BC\cos\theta}\right)}$

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{300\sin30^o}{200+300\cos30^o}\right)}$

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{300\times\dfrac{1}{2}}{200+300\times\dfrac{\sqrt3}{2}}\right)}$

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{150}{200+150\sqrt3}\right)}$

$\sf{\longrightarrow \alpha=\tan^{-1}\left(\dfrac{3}{4+3\sqrt3}\right)}$

$\sf{\longrightarrow\underline{\underline{\alpha=18.07^o}}}$

Hence the direction of city C relative to city A is 18.07° North of West.

Answer 2

Answer:

sorry............................................

Explanation: