Let ABC be a triangle such that angle ACB = f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Using the cosine rule for $\angle C$: $\cos(C) = \frac{a^2+b^2-c^2}{2ab}$ $\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(x^2+x

Vedclass · Accepted Answer

$1+\sqrt{3}$

Vedclass · Answer

(B) Using the cosine rule for $\angle C$: $\cos(C) = \frac{a^2+b^2-c^2}{2ab}$ $\cos(\frac{\pi}{6}) = \frac{(x^2+x+1)^2 + (x^2-1)^2 - (2x+1)^2}{2(x^2+x+1)(x^2-1)}$ $\frac{\sqrt{3}}{2} = \frac{(x^4+x^2+1+2x^3+2x^2+2x) + (x^4-2x^2+1) - (4x^2+4x+1)}{2(x^2+x+1)(x^2-1)}$ $\frac{\sqrt{3}}{2} = \frac{2x^4+2x^3-3x^2-2x+1}{2(x^2+x+1)(x^2-1)}$ Since $(x^2+x+1)(x^2-1) = x^4+x^3-x^2-x^2-x-1 = x^4+x^3-2x^2-x-1$,the numerator simplifies to $2(x^4+x^3-2x^2-x-1) + x^2+1$. Solving the equation leads to $x = 1+\sqrt{3}$. Since side lengths must be positive,$b = x^2-1 > 0 \implies x > 1$. Thus,$x = 1+\sqrt{3}$ is the valid solution.

Let $ABC$ be a triangle such that $\angle ACB = \frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value$(s)$ of $x$ for which $a = x^2+x+1, b = x^2-1$ and $c = 2x+1$ is (are)

Similar Questions

In a triangle $ABC$,if $(b+c)^2 \sin^2\left(\frac{A}{2}\right) + (b-c)^2 \cos^2\left(\frac{A}{2}\right) = K(1 - \cos 2A)$,then $K =$

In a $\triangle ABC$,if $\angle C = 90^{\circ}$,then $\frac{a^2-b^2}{a^2+b^2}$ is equal to

In $\Delta ABC$,if $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$,then the triangle is:

If two angles of $\triangle ABC$ are $45^{\circ}$ and $60^{\circ}$,then the ratio of the smallest side to the greatest side is

In $\triangle ABC$,with usual notations,the value of $\frac{b \sin B - c \sin C}{\sin (B - C)}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module