The number of real numbers x such that there | vedclass

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

Question

(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.For an isosceles triangle,at

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Let the angles of the triangle be $\alpha, \beta, \gamma$. We are given two angles as $A = 2x + 7$ and $B = 7x + 10$.For an isosceles triangle,at least two angles must be equal. We consider three cases:Case $1$: $A = B$$2x + 7 = 7x + 10$ $\Rightarrow 5x = -3$ $\Rightarrow x = -0.6$.The angles are $A = 5.8^\circ, B = 5.8^\circ, C = 180 - 11.6 = 168.4^\circ$. This is a valid triangle.Case $2$: $A = C$Since $A + B + C = 180^\circ$,we have $2A + B = 180^\circ$.$2(2x + 7) + (7x + 10) = 180$ $\Rightarrow 4x + 14 + 7x + 10 = 180$ $\Rightarrow 11x = 156$ $\Rightarrow x = \frac{156}{11} \approx 14.18$.The angles are $A = 35.36^\circ, B = 109.26^\circ, C = 35.36^\circ$. This is a valid triangle.Case $3$: $B = C$Since $A + B + C = 180^\circ$,we have $A + 2B = 180^\circ$.$(2x + 7) + 2(7x + 10) = 180$ $\Rightarrow 2x + 7 + 14x + 20 = 180$ $\Rightarrow 16x = 153$ $\Rightarrow x = \frac{153}{16} = 9.5625$.The angles are $A = 26.125^\circ, B = 76.9375^\circ, C = 76.9375^\circ$. This is a valid triangle.Since all three cases yield valid real values for $x$,the total number of such real numbers is $3$.

The number of real numbers $x$ such that there exists an isosceles triangle having two of its angles measured in degrees equal to $2x + 7$ and $7x + 10$ is:

Similar Questions

$\alpha$ is a root of the equation $\frac{x-1}{\sqrt{2x^2-5x+2}} = \frac{41}{60}$. If $-\frac{1}{2} < \alpha < 0$,then $\alpha = $

If $A$ and $G$ are arithmetic and geometric means and ${x^2} - 2Ax + {G^2} = 0$,then

If $x$ is real,then the minimum value of $y = \frac{x^2-x+1}{x^2+x+1}$ is

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+2x^2-x-2=0$,then $\alpha^6+\beta^6+\gamma^6=$

If $a, b, c$ are distinct and the roots of $(b-c)x^2 + (c-a)x + (a-b) = 0$ are equal,then $a, b$ and $c$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module