Question

To find the distance between a point X and an inaccessible point​ Z, a line segment XY is constructed. Measurements show that XYequals946 ​m, angle XYZequals12 degrees 36 prime​, and angle YZXequals99 degrees 42 prime. Find the distance between X and Z to the nearest meter.

Answer 1

Answer:

$XZ=209\ m$

Step-by-step explanation:

step 1

Convert degrees and minutes to degrees

Remember that

$1^o=60'$

so

$m\angle XYZ=12^o36'=12^o+(36/60)=12.6^o$

$m\angle YZX=99^o42'=99^o+(42/60)=99.7^o$

step 2

Applying the law of sines

$\frac{XY}{sin(YZX)}=\frac{XZ}{sin(XYZ)}$

substitute the given values

$\frac{946}{sin(99.7^o)}=\frac{XZ}{sin(12.6^o)}$

solve for XZ

$XZ=\frac{946}{sin(99.7^o)}sin(12.6^o)=209\ m$