If 7 and 8 are the lengths of two sides of a | vedclass

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

Question

(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,

Vedclass · Accepted Answer

$21$

Vedclass · Answer

(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.Thus,the angles are $60^{\circ} - d, 60^{\circ}, 60^{\circ} + d$.The sides opposite to these angles are $a, b, c$. Given two sides are $7$ and $8$. Let the sides be $a, 7, 8$ where $a$ is the smallest side.Using the Law of Cosines for the angle $60^{\circ}$:Case $1$: If the angle $60^{\circ}$ is between sides $7$ and $8$,then $a^2 = 7^2 + 8^2 - 2(7)(8) \cos 60^{\circ} = 49 + 64 - 56 = 57$. This gives $a = \sqrt{57} \approx 7.55$,which is not smaller than $7$.Case $2$: If the angle $60^{\circ}$ is opposite to side $7$,then $8^2 = a^2 + 7^2 - 2(a)(7) \cos 60^{\circ}$ $\Rightarrow 64 = a^2 + 49 - 7a$ $\Rightarrow a^2 - 7a - 15 = 0$. This yields $a = \frac{7 \pm \sqrt{49 + 60}}{2} = \frac{7 \pm \sqrt{109}}{2}$. Only the positive value is valid,$a \approx 8.72 > 7$.Case $3$: If the angle $60^{\circ}$ is opposite to side $8$,then $7^2 = a^2 + 8^2 - 2(a)(8) \cos 60^{\circ}$ $\Rightarrow 49 = a^2 + 64 - 8a$ $\Rightarrow a^2 - 8a + 15 = 0$.Solving $a^2 - 8a + 15 = 0$ gives $(a - 3)(a - 5) = 0$,so $a = 3$ or $a = 5$.Since $a$ is the smallest side,both $3 Thus,$a_1 = 3$ and $a_2 = 5$.Finally,$2 a_1 + 3 a_2 = 2(3) + 3(5) = 6 + 15 = 21$.

If $7$ and $8$ are the lengths of two sides of a triangle and '$a$' is the length of its smallest side. The angles of the triangle are in $AP$ and '$a$' has two values $a_1$ and $a_2$ satisfying this condition. If $a_1 < a_2$ then $2 a_1 + 3 a_2 =$

Similar Questions

In $\triangle ABC$,if $a, b, c$ are its sides and $\angle C = 60^{\circ}$,find the value of $\frac{a}{b+c} + \frac{b}{c+a}$.

Evaluate $\cot \left( \frac{A + B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right)$ in terms of sides $a$ and $b$.

The perimeter of a $\triangle ABC$ is $6$ times the arithmetic mean of the values of the sine of its angles. If its side $BC$ is of unit length,then $\angle A=$

In a triangle,the lengths of the two larger sides are $10 \, cm$ and $9 \, cm$ respectively. If the angles of the triangle are in $A.P.$,then the length of the third side in $cm$ can be

In a $\triangle ABC$,$2(bc \cos A + ca \cos B + ab \cos C) = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module