Trigonometric EquationHard
Question
There exists a triangle ABC satisfying the conditions
Options
A.b sin A = a , A < 
B.b sin A > a, A >
C.b sin A > a, A <
D.b sin A < a, A < , b > a
, b > aSolution
The sine formula is 
⇒ a sin B = b sin A
(a) : b sin A = a ⇒ a sin B = a ⇒ B =
Since, ∠A < therefore, the triangle is possible.
(b) and (c) : bsin A > = a ⇒ asin B > a ⇒ sin B > 1
∴ ᐃ ABC is not possible.
(d) : bsin A < a ⇒ asin B < a
⇒ sin B < 1 ⇒ ∠B exists
Now, b > a ⇒ B > A since A <
∴ The triamgle is possible.
⇒ a sin B = b sin A (a) : b sin A = a ⇒ a sin B = a ⇒ B =
Since, ∠A <
therefore, the triangle is possible. (b) and (c) : bsin A > = a ⇒ asin B > a ⇒ sin B > 1
∴ ᐃ ABC is not possible.
(d) : bsin A < a ⇒ asin B < a
⇒ sin B < 1 ⇒ ∠B exists
Now, b > a ⇒ B > A since A <
∴ The triamgle is possible.
Create a free account to view solution
View Solution FreeMore Trigonometric Equation Questions
If tan a θ - tan b θ = 0, then the value of θ for a series in -...The value of θ satisfying 3cos2 θ − 2√3 sin θ cos θ − 3 sin2 θ = 0 are-...The number of solutions of the equation sin(ex) = 5x + 5-x is...(t sin(2010t) + 2009t + 1006) dx is equal to -...If 4sin2θ + 4cosec2θ = 17, then the value of tan θ is -...