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Given: BD⎯⎯⎯⎯⎯ is an altitude of △ABC.Prove: sinAa=sinCcTriangle A B C with an altitude B D where D is on side A C. side A C is also labeled as small b. Side A B is also labeled as small c. Side B C is also labeled as small a. Altitude B D is labeled as small h. Select from the drop-down menus to correctly complete the proof.Statement ReasonBD⎯⎯⎯⎯⎯ is an altitude of △ABC. Given△ADB and △CDB are right triangles. Definition of right trianglesinA=hcand sinC=hacsinA=hand asinC=hMultiplication Property of EqualitycsinA=asinCcsinAac=asinCacsinAa=sinCcSimplify.

Given BD Is An Altitude Of ABCProve SinAasinCcTriangle A B C With An Altitude B D Where D Is On Side A C Side A C Is Also Labeled As Small B Side A B Is Also La class=

Sagot :

EXPLANATION :

From the problem, triangles ADB and CDB are right triangles.

In a right triangle, the sine function is the opposite side divided by the hypotenuse.

Referring to the figure,

sin A = h/c and sin C = h/a which are both definitions of sine ratio.

So the reason is Definition of sine ratio

By multiplication property of equality, use cross multiplication to remove the denominators

That's

c sinA = h

and

a sin C = h

Equating both equations in h :

c sin A = a sin C by the reason of Substitution Property of Equality

Diving both sides by "ac"

[tex]\frac{c\sin A}{ac}=\frac{a\sin C}{ac}[/tex]

by the reason of Division Property of Equality

Then :

[tex]\frac{\sin A}{a}=\frac{\sin C}{c}[/tex]

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