Question #6fc66

Aug 26, 2016

One cyclist traveling towards North-West and another towards South.So angle between them is $\theta = {\left(45 + 90\right)}^{\circ} = {135}^{\circ}$

Let the starting point of journey of two cyclists be O. After t hr first cyclist reaches at A traveling @ 22km/hr and at the same time the second cyclist reaches at B traveling @ 14km/hr .

$\text{So in } \Delta A O B$

$\angle A O B = \theta = {135}^{\circ}$

$O A = 22 t \text{ km}$

$O B = 14 t \text{ km}$

$A B = 80 \text{ km}$

Applying triangle law

$A {B}^{2} = O {A}^{2} + O {B}^{2} - 2 O A \cdot O B \cos \theta$

$\implies {80}^{2} = {\left(22 t\right)}^{2} + {\left(14 t\right)}^{2} - 2 \cdot 22 t \cdot 14 t \cos {135}^{\circ}$

$\implies {t}^{2} \left(484 + 196 + 616 \cdot \cos {45}^{\circ}\right) = 6400$

$\implies {t}^{2} \left(484 + 196 + 616 \cdot \frac{1}{\sqrt{2}}\right) = 6400$

$\implies 1115.58 {t}^{2} = 6400$

$\implies {t}^{2} = \frac{6400}{1115.58}$

$\implies t = \sqrt{\frac{6400}{1115.58}} \approx 2.39 h r$