Trigonometrical Equations
medium

જો  કોઈ $0 < \alpha < \frac{\pi }{2}$ માટે ત્રિકોણ ની બાજુઓ $\sin \alpha ,\,\cos \alpha $ અને $\sqrt {1 + \sin \alpha \cos \alpha } $ આપેલ છે તો ત્રિકોણનો સૌથી મોટો ખૂણો......$^o$ મેળવો.

A

$150$

B

$90$

C

$120$

D

$60$

(AIEEE-2004)

Solution

(c) $a = \sin \alpha ,\,b = \cos \alpha ,\,c = \sqrt {1 + \sin \alpha \cos \alpha } $
$\cos C = \frac{{{a^2} + {b^2} – {c^2}}}{{2ab}}$=$\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha – (1 + \sin \alpha .\cos \alpha )}}{{2\sin \alpha .\cos \alpha }}$
= $\frac{{1 – 1 – \sin \alpha \cos \alpha }}{{2\sin \alpha \cos \alpha }}$
$\cos C = \frac{{ – 1}}{2} = \cos \frac{{2\pi }}{3}$==> $C = \frac{{2\pi }}{3} = 120^\circ $.

Standard 11
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.