You visited us 1 times! Enjoying our articles? Unlock Full Access!

Cosine Rule

If in a ABC...

Question

If in a $△ A B C, a^{2} {cos}^{2} A = b^{2} + c^{2},$ then

A < \frac{π}{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A

lies between

(\frac{π}{4}, \frac{π}{2})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

A > \frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

A = \frac{π}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $A > \frac{π}{2}$
$a^{2} {cos}^{2} A = b^{2} + c^{2}$ ............... $(1)$
$∵ cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$= \frac{a^{2} {cos}^{2} A - a^{2}}{2 b c}$ from $(1)$
$= \frac{- a^{2}}{2 b c} (1 - {cos}^{2} A) = \frac{- a^{2}}{2 b c} {sin}^{2} A < 0$
$\Rightarrow cos A < 0$
or $A > \frac{π}{2}$

Suggest Corrections

Similar questions

(a + \sqrt{2} b cos x) (a - \sqrt{2} b cos y) = a^{2} - b^{2}

a > b > 0

\frac{d x}{d y}

(\frac{π}{4}, \frac{π}{4})

z = \frac{1 + 2 i cos θ}{4 + 3 i sin θ}

θ \in (0, \frac{π}{2})

8 cos x (cos (\frac{π}{6} + x) cos (\frac{π}{6} - x) - \frac{1}{2}) = 1

[0, π]

k π,

k

\int \sqrt{\frac{cos x - {cos}^{3} x}{1 - {cos}^{3} x}} d x

x \in (0, \frac{π}{2})

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{1 - tan x}{4 x - π}, & x \neq \frac{π}{4} λ, & x = \frac{π}{4} \end{matrix}

x ϵ [0, \frac{π}{2})

f (x)

[0, \frac{π}{2})

λ

Join BYJU'S Learning Program