In a triangle ABC,∠$$C=$$π$$2$$. If $$tan$$⁡$$A2$$ and $$tan$$⁡$$B2$$ are the roots of the equation $$ax2+bx+c=0(a$$≠$$0)$$,then the value of $$a+bc$$ (where$$ a, b, c are sides of ∆ opposite to angles A, B, C, $$respectively) is

$$tan$$⁡$$A2+$$$$tan$$⁡$$B2=$$−batan⁡$$A2$$×$$tan$$⁡$$B2=caA+B=90$$∘ or $$A+B2=45$$∘⇒$$tan$$⁡$$A+B2=1=$$−$$ba1$$−caor $$1$$−$$ca=$$−baor $$a+b=cor$$ $$a+bc=1$$

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Question

In a triangle $ABC, ∠ C = \frac{π}{2} . If \tan (\frac{A}{2})$ and $\tan (\frac{B}{2})$ are the roots of the equation ${ax}^{2} + bx + c = 0 (a \neq 0)$ ,then the value of $\frac{a + b}{c}$
(where a, b, c are sides of $∆$ opposite to angles A, B, C, respectively) is

Moderate

Solution

$\begin{matrix} \tan (\frac{A}{2}) + \tan (\frac{B}{2}) = - \frac{b}{a} \\ \tan (\frac{A}{2}) \times \tan (\frac{B}{2}) = \frac{c}{a} \\ A + B = 90^{\circ} or \frac{A + B}{2} = 45^{\circ} \\ \Rightarrow \tan (\frac{A + B}{2}) = 1 = \frac{- \frac{b}{a}}{1 - \frac{c}{a}} \end{matrix}$
$\begin{array}{l} or 1 - \frac{c}{a} = - \frac{b}{a} \\ or a + b = c \\ or \frac{a + b}{c} = 1 \end{array}$

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Compound angles

Remember concepts with our Masterclasses.

In a triangle ABC,∠C=π2. If tan⁡A2 and tan⁡B2 are the roots of the equation ax2+bx+c=0(a≠0),then the value of a+bc (where a, b, c are sides of ∆ opposite to angles A, B, C, respectively) is

tan⁡A2+tan⁡B2=−batan⁡A2×tan⁡B2=caA+B=90∘ or A+B2=45∘⇒tan⁡A+B2=1=−ba1−caor 1−ca=−baor a+b=cor a+bc=1

Ready to Test Your Skills?

free study material

In a triangle $ABC, ∠ C = \frac{π}{2} . If \tan (\frac{A}{2})$ and $\tan (\frac{B}{2})$ are the roots of the equation ${ax}^{2} + bx + c = 0 (a \neq 0)$ ,then the value of $\frac{a + b}{c}$
(where a, b, c are sides of $∆$ opposite to angles A, B, C, respectively) is