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AMC 10 USA Math Olympiad

Problem on Trigonometry | SMO, 2008 | Problem – 22

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry. You may use sequential hints to solve the problem.

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2008 based on Trigonometry.

Problem on Trigonometry | SMO, 2008 |Problem 22

Find the value of \(\frac {tan 40^\circ tan 60^\circ tan 80^\circ}{tan40^\circ + tan 60^\circ + tan 80^\circ}\)

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Key Concepts


Trigonometry

Tan Rule

Check the Answer


But try the problem first…

Answer: 1

Source
Suggested Reading

Singapore Mathematical Olympiad, 2008

Challenges and Thrills – Pre College Mathematics

Try with Hints


First hint

If you got stuck in this sum how to get started you can start by consider a general case where \(40^\circ = A\) , \(60^\circ = B\) and \(80^\circ = C\).

So , A+B+C = \( 180 ^\circ\)

\( A+B = 180^\circ – C\)

(tan (A+B) = tan \(180^\circ – C)\)……………………….(1)

Now try to implement the basic formula and try to do this sum………………

Second Hint

In this we can continue from the last hint:

the formula of tan (A + B) = \(\frac {tan A + tan B}{1- tan A . tan B}\)

From the equation (1) …….

tan (A+B) = tan (180 – C)

\(\frac {tan A + tan B}{1- tan A . tan B} = tan (180^\circ – c)\)

(frac {tan A + tan B}{1- tan A . tan B} = -tan C )

Now just rearrange this expression and you will get the final answer……………..

Final Step

Here is the final solution:

{tan A + tan B} = -tan C {1- tan A . tan B}

tan A + tan B = – tan C + tan C tan A tan B

tan A + tan B + tan C = tan A tan B tan C

\(\frac {tan A tan B tan C}{tan A + tan B + tan C} = 1\)

Which is the given question. It can be a proof also……………….

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