Determine the largest angle if the angles of a triangle are in the ratio 3 : 4 : 5.
75°
The correct answer is 75°.
Obtuse-angled triangle: One angle is greater than $90^\circ$. The ratio 3:4:5 indicates that the angle measures increase proportionally. The largest part of the ratio (5) corresponds to the largest angle, and the smallest part (3) corresponds to the smallest angle. This relationship holds true for any set of positive numbers in a ratio representing angles.
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
The value of cot (-315°) is :
The value of \(\frac{\tan^2 (22^\circ - \theta)-\tan (\theta + 68^\circ)\ -\ \text{cosec}^2 (68^\circ + \theta)+\cot (22^\circ - \theta)} {3 (\cot^2 52^\circ - \sec^2 38^\circ)+ 2 (\text{cosec}^2 \ 28^\circ - \tan^2 62^\circ)}\) is:
The value of \(\frac{{{{\sec }^2}60^\circ {{\cos }^2}45^\circ \, + \,{\rm{cose}}{{\rm{c}}^2}30^\circ }}{{{{\sec }^2}45^\circ - \tan 45^\circ }}\) is:
The value of \(\sin^2\frac{2\pi}{3}+\cos^2\frac{5\pi}{6}-\tan^2\frac{3\pi}{4}\) is :
sinθ + cosec θ = 2, then the value of sin 99 θ + cosec 99 θ is:
If cos θ = \(\frac{5}{13}\) , what is the value of cot θ ?
The value of the expression \(\rm \frac{4\sin^230^{\circ}+\cos^260^{\circ}-\tan^245^{\circ}}{2\sin60^{\circ}\cos30^{\circ}-\tan45^{\circ}}\) is:
The general solution of the equation \(\tan 3x + \cot (2x + \frac{\pi }{3}) = 0\) is:
If tan 15θ = cot 15θ (0° < θ < 10°) , then the value of θ is:
Which of the following values of A and B satisfies, sin(A + B) = sin A + sin B, where
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
The value of 4cos\(\left( {\frac{\pi }{6}\, - \,\alpha } \right)\) sin\(\left( {\frac{\pi }{3}\, - \,\alpha } \right)\) is equal to:
1 + tan 15° cot 75° is equal to:
If tan θ = 1/√5, find the value of cosec2θ – sec2θ.