In triangle ABC,the sum of the lengths of two | vedclass

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

Question

(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$. Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$: $(a + b)^2

Vedclass · Accepted Answer

$\frac{c}{\sqrt{3}}$

Vedclass · Answer

(A) Let the two sides be $a$ and $b$. We are given $a + b = x$ and $ab = y$. Given the condition $x^2 - c^2 = y$,we substitute $x = a + b$: $(a + b)^2 - c^2 = y$ $a^2 + b^2 + 2ab - c^2 = y$ Since $ab = y$,we have $a^2 + b^2 + 2y - c^2 = y$,which simplifies to $a^2 + b^2 - c^2 = -y$. By the Law of Cosines,$c^2 = a^2 + b^2 - 2ab \cos C$. Comparing $a^2 + b^2 - c^2 = 2ab \cos C$ with $a^2 + b^2 - c^2 = -y$ and $ab = y$,we get $2y \cos C = -y$,so $\cos C = -\frac{1}{2}$. Thus,$C = 120^\circ$. The circumradius $R$ is given by $R = \frac{c}{2 \sin C}$. $R = \frac{c}{2 \sin 120^\circ} = \frac{c}{2 (\sqrt{3}/2)} = \frac{c}{\sqrt{3}}$.

In $\triangle ABC$,the sum of the lengths of two sides is $x$ and the product of those lengths is $y$. If $c$ is the length of the third side and $x^2 - c^2 = y$,then the circumradius of the triangle is

Similar Questions

In triangle $ABC$,if $r_1=36, r_2=18$ and $r_3=12$,then $a+b=$

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

If the median of a $\triangle ABC$ through $A$ is perpendicular to $AC$,then $\frac{\tan A}{\tan C}=$

In a $\Delta ABC,$ if $\frac{\sin A}{\sin C} = \frac{\sin (A - B)}{\sin (B - C)},$ then $a^2, b^2, c^2$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module