Question

In a triangle sin A ,sin B ,sin C are in A P then...
a) altitudes are in AP
B)the al;titudes are in HP
C)the altitudes are in G P
D)none of these

Answer 1

@phi, @myininaya

Answer 2

@amistre64

Answer 3

I would guess B. Harmonic progression. since due to Sine law, inverse relations build up between the sides and sine values. I haven't solved it properly, but I ll let you know.

Answer 4

@Mani_Jha

Answer 5

@satellite73

Answer 6

@KingGeorge

Answer 7

ok at first glance i am going to say arithmetic. let me see if can draw a picture

Answer 8

nope that is not working out for me

Answer 9

hmm ..these days my qns makes people here to think more

Answer 10

:P

Answer 11

yes it is a lot of thinking. so far i have thought this
\[h_1=b\sin(A)\]
\[h_2=c\sin(B) = c(\sin(A)+d)=c\sin(A)+cd\]
\[h_3=a\sin(C)=a(\sin(A)+2d)=a\sin(A)+2ad\]

Answer 12

and
\[\frac{c}{\sin(C)}=\frac{a}{\sin(A)}\]
\[\frac{c}{\sin(A)+2d}=\frac{a}{\sin(A)}\] so
\[c=\frac{a(\sin(A)+2d}{\sin(A)}\] and now maybe we can write \(h_2\) in term so only of "a" and \(\sin(A)\) and "d"

Answer 13

@nikvist

Answer 14

\[h_2=\frac{a(\sin(A)+2d)}{\sin(A)}(\sin(A)+d)\]}

Answer 15

\[h_3=a(\sin(A)+2d)\] so it looks like \(h_2\) is a multiplye of \(h_3\)

Answer 16

similarly
\[b=\frac{a(\sin(A)+d)}{\sin(A)}\] so
\[h_1=\frac{a(\sin(A)+d)}{\sin(A)}\times \sin(A)=a(\sin(A)+d)\]

Answer 17

did i screw this up yes?

Answer 18

*yet

Answer 19

i think there is an easier way

Answer 20

certainly a more systematic way i am sure

Answer 21

you have
\[\sin(A), \sin(B)=\sin(A)+d, \sin(C)=\sin(A)+2d\] then they are in arithemtic progression
now you have to write the altitudes in term of the sides and the sines

Answer 22

@FoolForMath , @Hero

Answer 23

but the only way i can see to do it is write
\[h_1=b\sin(A)\] and then solve for "b" in terms of "a" and \(\sin(A)\)

Answer 24

@Sarkar

Answer 25

I am trying to find those three angles programmatically ... lol

Answer 26

@AravindG I think you should give them a medal, don't be stingy.... at least they're trying to help you!

Answer 27

improved my code by at lot still no luck ... anyone here good at programming??

Answer 28

@kreshnik u see i was having dinner .. no qn i have left without giving medals!!