In triangle ABC, A-D-B, CD=6, BD=9 and BC=12. If angle CAB=BCD then what is the perimeter of triangle ADC ?

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In triangle ABC, A-D-B, CD=6, BD=9 and BC=12. If angle CAB=BCD then what is the perimeter of triangle ADC ?

2006-10-26 21:44:16 · 3 answers · asked by ankur s 1 in Science & Mathematics ➔ Mathematics

The answer to the question is given to be 36, but how?

2006-10-27 05:03:20 · update #1

3 answers

using laws of sine and laws of cosine,

the answer is 21.

for use law of cosine to find angle DCB and BDC

then use 180 - BDC to find ADC

use 6/sin CAB = x /sinADC
find x

use 180 - CAD - ADC to find ACD

then use 6/sinCAD = y/sin ACD
find y

perimeter is x + y + 6 = 21 units

2006-10-26 22:04:41 · answer #1 · answered by sumone^^ 3 · 0⤊ 0⤋

Yes, ABC is the 1st 3 leters of our alphabet!

2006-10-26 21:55:37 · answer #2 · answered by Anonymous · 0⤊ 0⤋

27 i think.

Boy this one had me scratching my head, try a diagram next time.

2006-10-26 21:54:02 · answer #3 · answered by Yvonne Mystic 4 · 0⤊ 0⤋