Question

HELP PLEASE 98 points

Given: ​ BD¯¯¯¯¯ ​ is an altitude of △ABC .


Prove: sinAa=sinCc

Triangle 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.




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Statement Reason

​ BD¯¯¯¯¯ ​ is an altitude of △ABC . Given

​ △ABD ​ and ​ △CBD ​ are right triangles. Definition of right triangle

sinA=hc and sinC=ha

csinA=h and asinC=h

csinA=asinC

csinAac=asinCac Division Property of Equality

sinAa=sinCc Simplify.

Answer 1

Answer:

Given: ​ BD ​is an altitude of △ABC .

Prove: sinA/a=sinC/c

Triangle ABC with an altitude BD where D is on side AC. Side AC is also labeled as small b. Side AB is also labeled as small c. Side BC is also labeled as small a. Altitude BD is labeled as small h.

Statement Reason

​ BD is an altitude of △ABC .

Given △ABD ​ and ​ △CBD ​ are right triangles. (Definition of right triangle)

sinA=h/c and sinC=h/a

Cross multiplying, we have

csinA=h and asinC=h

(If a=b and a=c, then b=c)

csinA=asinC

csinA/ac=asinC/ac (Division Property of Equality)

sinA/a=sinC/c

This rule is known as the Sine Rule.

Answer 2

Answer:

For the people on Edge the answers are:

a/f

cf

c/b

cf + ec

c