the Red Line in the figure is an altitude of triangle hjl. Using right angle trig and properties of equality, y sin L = x = z sin H, write the Law of Sines for this triangle

the Red Line in the figure is an altitude of triangle hjl Using right angle trig and properties of equality y sin L x z sin H write the Law of Sines for this tr class=

Respuesta :

Answer:

[tex]\Large \boxed{\mathrm{\bold{C}}}[/tex]

Step-by-step explanation:

[tex]\sf \displaystyle sin(\theta) =\frac{opposite}{hypotenuse }[/tex]

[tex]\sf \displaystyle sin(L) =\frac{x}{y}[/tex]

[tex]\sf \displaystyle sin(H) =\frac{x}{z}[/tex]

[tex]\displaystyle \sf \frac{sinL}{z} =\frac{sinH }{y}[/tex]

[tex]\displaystyle \sf \frac{\frac{x}{y} }{z} =\frac{\frac{x}{z} }{y}[/tex]

Simplifying the expression.

[tex]\sf \displaystyle \frac{x}{yz } = \frac{x}{yz} \ (true)[/tex]

[tex]\displaystyle \sf \frac{z}{sinL} =\frac{y}{sinH}[/tex]

[tex]\displaystyle \sf \frac{z}{\frac{x}{y}} =\frac{y}{\frac{x}{z}}[/tex]

Multiplying both sides by x.

[tex]\sf yz=yz \ (true)[/tex]

Answer:

d

Step-by-step explanation:

The law of Sines applied to a Δ ABC is

[tex]\frac{a}{sinA}[/tex] = [tex]\frac{b}{sinB}[/tex] = [tex]\frac{c}{sinC}[/tex]

Using this in Δ HJL

[tex]\frac{z}{sinL}[/tex] = [tex]\frac{y}{sinH}[/tex] → (b)

Which may be expressed in reciprocal form as

[tex]\frac{sinL}{z}[/tex] = [tex]\frac{sinH}{y}[/tex] → (a)

Thus the law of Sines can be expressed as either (a) or (b)

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