If the data given to construct a triangle ABC are a = 5, b = 7, \(\sin A = \frac{3}{4}\), then it is possible to construct
no triangle
Concept:
Sine (sin θ) is a function that represents the shape of a right triangle. Looking from a vertex with angle θ, sinθ in a right triangle is the ratio of the opposite side to the hypotenuse. Additionally, we know that in any right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. Therefore, we need to find the third side of the triangle.
Calculation:
a = 5, b = 7, sin A = 3/4
We know that,
sin2A + cos2A = 1
⇒ cos A = \(\sqrt{1 - sin^2 A}\)
⇒ cos A = \(\sqrt{1 - \frac{9}{16}}\)
⇒ cos A = \(\sqrt{\frac{7}{16}}\)
⇒ cos A = \(\frac{\sqrt{7}}{4}\)
Formula: Law of Cosines
\(cos A = \frac{b^2 + c^2 - a^2}{2 bc}\) , where c = length of side c, a = length of side a, b = length of side b, and A = the angle opposite to c
⇒ \(\frac{\sqrt{7}}{4}\) = \(\frac{49 + c^2 - 25}{2 (7)c}\)
⇒ 2c2 + 48 - \(7\sqrt{7}\) c = 0
To determine the nature of the roots of the quadratic equation, we need D = \(\sqrt{b^2-4ac}\)
⇒ D = 343 - 384 < 0
Thus, the roots are not real. Therefore, no triangle is possible since the sides of a triangle cannot be imaginary.
In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are
Which of the following measures can form a triangle?
Which of the following cannot be the sides of a triangle?
If in a triangle ABC, \(\frac{{2\cos A}}{a} + \frac{{\cos B}}{b} + \frac{{2\cos C}}{c} = \frac{a}{{bc}} + \frac{b}{{ca}}\) then the value of the angle A is
In a triangle ABC, sec A (sin B cos C + cos B sin C) equals: