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Let A,B,C be three angles such that A={} {4} and B C=p. Find

Question

Let A,B,C be three angles such that $A = \frac{π}{4}$ and $tan B tan C = p$ . Find all possible values of p such that A,B,C are the angles of a triangle.

p < 0

p > (2 + 1)^{2}

p > 0

p < (2 + 1)^{2}

p > 0

p < (2 - 1)^{2}

p < 0

p > (2 - 1)^{2}

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Lucas Discussed

Let A,B,C be three angles such that

A = \frac{π}{4}

and

tan B tan C = p

. Find all possible values of p such that A,B,C are the angles of a triangle.

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

A + B + C = π \Rightarrow B + C = π - \frac{π}{4} = \frac{3 π}{4}

...(1)

(∵ A = \frac{π}{4})

tan B . tan C = p \Rightarrow \frac{sin B . sin C}{cos B . cos C} = \frac{p}{1}

\Rightarrow \frac{sin B . sin C + cos B . cos C}{sin B . sin C - cos B . cos C} = \frac{p + 1}{p - 1}

\Rightarrow \frac{cos ( A - C )}{cos ( B + C )} = \frac{1 + p}{1 - p}

\Rightarrow cos (B - C) = \frac{( 1 + p )}{2 ( 1 - p )}

...(2)

(∵ B + C = \frac{3 π}{4})

Since

B

C

can vary from

0

\frac{3 π}{4}

0 \leq B - C < \frac{3 π}{4} \Rightarrow - \frac{1}{2} < cos (B - C) \leq 1

...(3)

From (2) and (3), we get

- \frac{1}{2} < \frac{1 + p}{2 ( p - 1 )} \leq 1

\Rightarrow - \frac{1}{2} < \frac{1 + p}{2 ( p - 1 )}

and

\frac{1 + p}{2 ( p - 1 )} \leq 1

\Rightarrow \frac{1 + p}{p - 1} + 1 \geq 0

and

\frac{1 + p - 2 p + 2}{2 ( p - 1 )} \leq 0

\Rightarrow \frac{2 p}{p - 1} \geq 0

and

\frac{( 1 - 2 ) ( p - \frac{1 + 2}{1 - 2} )}{2 ( p - 1 )} \leq 0

\Rightarrow \frac{2 p}{p - 1} > 0

and

\frac{( p - ( 2 + 1 ) ^{2} )}{( p - 1 )} \geq 0

\Rightarrow (p < 0 or p > 1)

and

(p < 1 or p > (2 + 1)^{2})

p < 0

p \geq (2 + 1)^{2}

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Question Text	Let A,B,C be three angles such that $A = \frac{π}{4}$ and $tan B tan C = p$ . Find all possible values of p such that A,B,C are the angles of a triangle.
Answer Type	Text solution:1
Upvotes	150