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# If in a triangle ABC, $\tan A+\tan B+ \tan C=6$ and $\tan A\tan B=2$ then the triangle is:A. Right angledB. Obtuse angledC. Acute angledD. Isosceles

Last updated date: 17th Sep 2024
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Hint: In this question, we are given triangle ABC, therefore angles of $\Delta ABC$ are $\angle A,\angle B\text{ and }\angle C$. We are given $\tan A+\tan B+ \tan C=6$ and $\tan A\tan B=2$. We need to find the type of triangle.
For this, we need to find ... using values of tanA, tanB, tanC found from the given two conditions. We will use the following properties for solving this sum:
1. Sum of angles of the triangle is ${{180}^{\circ }}$.
2. $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.
3. $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta$.
4. If tanA, tanB, tanC are positive, then the value of tangent will lie between 0 and $\dfrac{\pi }{2},\pi \text{ and }\dfrac{3\pi }{2}$.

Complete step-by-step solution
Here, we are given triangle ABC therefore, angles of $\Delta ABC$ are $\angle A,\angle B\text{ and }\angle C$. We are given that, $\tan A+\tan B+ \tan C=6$ and $\tan A\tan B=2$.
Let us find the value of tanA, tanB, tanC. As we know, the sum of angles of a triangle is ${{180}^{\circ }}$. Therefore, in $\Delta ABC$, $\angle A+ \angle B+ \angle C={{180}^{\circ }}$
We can also say $A+B+C={{180}^{\circ }}\Rightarrow A+B={{180}^{\circ }}-C$.
Using tan on both sides we get: $\tan \left( A+B \right)=\tan \left( {{180}^{\circ }}-C \right)$.
We know that, $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta$ then we get: $\tan \left( A+B \right)=-\tan C$.
As we know that $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ so we get: $\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C$.
Cross multiplying we get: $\tan A+\tan B=-\tan C\left( 1-\tan A\tan B \right)\Rightarrow \tan A+\tan B=-\tan C+tanA\tan B\tan C$
We know that, $\tan A\tan B=2$ so we get:
\begin{align} & \Rightarrow \tan A+\tan B=-\tan C+\left( 2 \right)\tan C \\ & \Rightarrow \tan A+\tan B=-\tan C+2\tan C \\ & \Rightarrow \tan A+\tan B=\tan C \\ \end{align}
Adding tanC on both sides we get:
$\Rightarrow \tan A+\tan B+\tan C=\tan C+\tan C$
As we know, $\tan A+\tan B+\tan C=6$ so we get: $6=2\tan C$.
Dividing both sides by 2, we get $\tan C=3$.
Also we get: $\tan A+\tan B=3$.
Now we have $\tan A+\tan B=3$ and $\tan A\tan B=2$. So values of tanA and tanB should be 2 and 1.
So, either tanA = 2 and tanB = 1 or tanA = 1 and tanB = 2.
Now, we have found that tanA, tanB, tanC are all positive, therefore, A, B, C will lie between 0 and ${{90}^{\circ }}$ or ${{180}^{\circ }}\text{ and }{{270}^{\circ }}$. Since A, B, C is angles of the triangle and cannot be greater than ${{180}^{\circ }}$ so values of A, B, C will lie between 0 and ${{90}^{\circ }}$. Therefore, angles A, B, C are acute angles.
Hence, $\Delta ABC$ is an acute-angled triangle.
Therefore, option C is the correct answer.

Note: Students should take care of the signs while applying trigonometric formula such as $\tan \left( {{180}^{\circ }}-\theta \right)=-\tan \theta$ Students should know trigonometric functions are positive in which quadrant. For example, in the first quadrant $\left( 0\text{ to }{{90}^{\circ }} \right)$ all trigonometric functions (sin,cos,tan) are positive. In the second quadrant, $\left( {{90}^{\circ }}\text{ to }{{180}^{\circ }} \right)$ sine is positive. In third quadrant, $\left( {{180}^{\circ }}\text{ to 27}{{0}^{\circ }} \right)$ tangent is positive and in fourth quadrant, $\left( {{270}^{\circ }}\text{ to 36}{{0}^{\circ }} \right)$ cosine is positive. If we have calculations, we can also find values of A, B, C by taking the inverse of the tangent.