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# If A, B, C are the angles of a triangle such that angle A is obtuse, then $\tan B \tan C$ will be less thanA) $\dfrac{1}{{\sqrt 3}}$B) $\dfrac{{\sqrt 3}}{2}$C) 1D) None of these

Last updated date: 20th Jun 2024
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We have been given that the angles A, B, C are angles of the same triangle therefore we can use the sum of angles property. Also, we have been given that Angle A is obtuse and therefore will lie between $\dfrac{\pi} {2}$ and $\pi$. We will use this to get a range of values for B and C and consequently tanBtanC.

Complete step-by-step solution
Figure:
Let us first draw a figure which represents the triangle and its angles A, B, and C, where $\angle$ A is obtuse.

Since, we have been given that A is an obtuse angle, we will express it mathematically as,
$\dfrac{\pi }{2} < A < \pi$ . . . . . . . . (1)
We know that A, B, C are angles of the same triangle. Therefore by the property Sum of angles of a triangle we get,
A+B+C = $\pi$
We need to get a range of values for B and C.
Angle A can be written as,
A = $\pi - \left( {B + C} \right)$
Let us substitute this in equation (1)
We get,
$\dfrac{\pi }{2} < \pi - (B + C) < \pi$
We want the angles B and C to be positive.
So we will multiply throughout the expression by” – “to keep the term (B+C) positive.
This will reverse the inequalities from “<” to “>”.
We get the following expression,
$- \dfrac{\pi }{2} > - \pi + (B + C) > - \pi$
To simplify further we will add $\pi$ throughout the expression. We get,
$\Rightarrow \dfrac{\pi }{2} > \left( {B + C} \right) > 0$
Therefore, we now know that B+C is less than$\dfrac{\pi }{2}$.
$\Rightarrow B + C < \dfrac{\pi }{2}$
$\Rightarrow B < \dfrac{\pi }{2} - C$
As we have to find the values of tanBtanC, we will apply tan on both sides of the inequality.
$\Rightarrow \tan (B) < \tan \left( {\dfrac{\pi }{2} - C} \right)$
Using the trigonometric property, $\tan \left( {\dfrac{\pi }{2} - \theta } \right) = \cot (\theta )$, we get
$\Rightarrow \tan B < \cot C$
We also know that, $\cot \theta = \dfrac{1}{{\tan \theta }}$. Therefore we get,
$\Rightarrow \tan B < \dfrac{1}{{\tan C}}$
We will now bring the terms $\tan B$ and $\tan C$ on the same side.
$\therefore \tan B\tan C < 1$
Hence, for the above-given question tanBtanC will be less than 1.
The correct answer choice is (C).

Note: We should keep in mind the two properties given in the question so that we keep the solution simple. Also, we should avoid opening the bracket initially as we might get confused. Keep in mind to keep (B+C) positive. We should also remember to get rid of the $\pi$ in the middle term by adding the expression with $\pi$.