which is correct?

Q : A,B,C are 3 angles of a triangle. A=Ï€/4 , tanB . tanC = p.
find all possible values of p.

my approach.
B+C= 3Ï€/4

so , tanB . tan(3Ï€/4 - B) = p.
tanB=x ,

x2 + x(1-p) + p =0 .

for real roots.
Δ≥0.

so , p2 -6p +1 ≥0
which implies p ≥3 + √8 , or p ≤ 3 - √8

but here in this book ,well the approach is different ,I'll give it later. we've got p < 0 , or p ≥ (√2 + 1 )2

9 Answers

62
Lokesh Verma ·

x2 + x(1-p) + p =0 .

should be

x2 + x(p-1) + p =0 .

Just check :)

62
Lokesh Verma ·

there's a caveat attached... the angle goes from pi/4 to pi only

so u will have to remove the other values of p :)

just see if that gives u the solution

1
voldy ·

which angle goes from pi/4 to pi?

62
Lokesh Verma ·

angle B

i guess that was a mistake i tshould have been pi/4 to 3pi/4

1
voldy ·

I didn't get you.

62
Lokesh Verma ·

so angle B will always lie between 0 and 3pi/4

1
voldy ·

then what happns? please xplain more . I'm dud in math . Better give your soln. atleast halfway.

62
Lokesh Verma ·

ok.. ok..

dont worry i will try to explain :)

I generally dont like giving the whole solution because i think that it inhibits the thinking process going in ur mind...

11
Shailesh ·

Hey srinath.. u are missing somthing.. I understand what nishant is trying to say..

It is not very tough.. i cant explain.. but i have got what nishant is trying to say.. just try one time.

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