plz. answer this question.....
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tan (theta) = AC/BC
tanФ = AC/CD = AC/(1/2BC) = 2AC/BC
tan(theta)/tanФ = (AC/BC)/(2AC/BC)
tan(theta)/tanФ =1/2
hence proved ;)
tanФ = AC/CD = AC/(1/2BC) = 2AC/BC
tan(theta)/tanФ = (AC/BC)/(2AC/BC)
tan(theta)/tanФ =1/2
hence proved ;)
Answered by
1
D is the midpoint of BC⇒CD=BD
BC=BD+CD=CD+CD=2CD
tanθ=AC/BC=AC/2CD
tanΦ=AC/CD=AC/CD
tanθ/tanΦ=(AC/2CD)/(AC/CD)=1/2
BC=BD+CD=CD+CD=2CD
tanθ=AC/BC=AC/2CD
tanΦ=AC/CD=AC/CD
tanθ/tanΦ=(AC/2CD)/(AC/CD)=1/2
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