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the bearing of two points x and y from z are 45° and 135° respectively . if |zx|=8cm and |zy|=6cm, find |xy|.

Sagot :

Answer:

[tex]|{\sf XY}| = 10\; {\rm cm}[/tex].

Step-by-step explanation:

Refer to the diagram attached. The dashed segment attached to [tex]\!{\sf Z}[/tex] points to the north. Rotating this segment clockwise with point [tex]{\sf Z}\!\![/tex] as the fixed center of rotation would eventually align this segment with the one between point [tex]\!\!{\sf Z}[/tex] and point [tex]\!\!{\sf X}[/tex]. The bearing of point [tex]{\sf X}[/tex] from point [tex]{\sf Z}[/tex] is the size of the angle between these two line segments when measured in the clockwise direction.

Subtract the bearing of [tex]{\sf Y}[/tex] from [tex]{\sf Z}[/tex] from the bearing of [tex]{\sf X}[/tex] from [tex]{\sf Z}[/tex] to find the measure of the angle [tex]\angle {\sf YZX}[/tex]:

[tex]\begin{aligned}\angle {\sf YZX} &= 135^{\circ} - 45^{\circ} \\ &= 90^{\circ}\end{aligned}[/tex].

Thus, triangle [tex]\triangle {\sf YZX}[/tex] is a right triangle ([tex]90^{\circ}[/tex]) with segment [tex]{\sf YX}[/tex] as the hypotenuse. It is given that [tex]|{\sf XZ}| = 6\; {\rm cm}[/tex] whereas [tex]|{\sf ZY}| = 6\; {\rm cm}[/tex]. Thus, by Pythagorean's Theorem:

[tex]\begin{aligned}|{\sf ZY}| &= \sqrt{|{\sf ZX}|^{2} + |{\sf ZY}|^{2}} \\ &= \sqrt{(8\; {\rm cm})^{2} + (6\; {\rm cm})^{2}} \\ &= 10\; {\rm cm}\end{aligned}[/tex].

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