Answer to Question #311781 in Trigonometry for Jay

Question

Am explorer walks 22km due North and then walks 45degrees south east for 47km.what’s the resultant displacement from the origin?

Answer 1

Suppose explorer start from origin $A$ to north $22Km$ to a point $B$.

Then Move $45\degree$ South East $47Km$ to reach a point $C$.

Join point $AC,$ Length of $AC$ would be displacement from origin

Now draw a line $DC$ from point $C$ that meet line $AB$ on point $D$

$\triangle ABC$ Given

$\angle B= 45 \degree$ $AB=22Km$

$BC=47 Km$

$Displacement =AC=?$

From $\triangle DBC$

$SinB=\frac{DC}{BC}$

$Sin45 \degree=\frac{DC}{47}$

$DC={BC}{Sin45 \degree}=47Sin45 \degree$

$DC=33.2340Km$

$\triangle DBC$

$TanB=\frac{DC}{BD}$

$Tan45 \degree=\frac{DC}{BD}$

$BD=DC=33.23Km$

$AD=BD-AB=33.23-22$

$AD=11.23Km$

$\triangle ADC$

${AC}^2={DC}^2 +{AD}^2$

${AC}^2={33.23}^2 +{11.23}^2$

${AC}^2=1230.3458$

$AC=\sqrt{1230.3458}$

$AC=35.0049 Km$

$$Displacement\boxed {AC=35.00 Km}$$