The angles A, B, C of a triangle ABC are in A | vedclass

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

Question

(C) Given that the angles $A, B, C$ of $	riangle ABC$ are in $AP$,we have $A+C = 2B$.Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,which impl

Vedclass · Accepted Answer

$\sqrt{43}$

Vedclass · Answer

(C) Given that the angles $A, B, C$ of $	riangle ABC$ are in $AP$,we have $A+C = 2B$.Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.Using the cosine rule in $	riangle ABC$:$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 	imes AB 	imes BC}$Substituting the given values $AB=6, BC=7, B=60^{\circ}$ and letting $AC=x$:$\cos 60^{\circ} = \frac{6^2 + 7^2 - x^2}{2 	imes 6 	imes 7}$$\frac{1}{2} = \frac{36 + 49 - x^2}{84}$$42 = 85 - x^2$$x^2 = 85 - 42 = 43$$x = \sqrt{43}$Therefore,$AC = \sqrt{43}$.

The angles $A, B, C$ of a triangle $ABC$ are in $AP$. If $AB=6$ and $BC=7$,then $AC=$

Similar Questions

In a $\triangle ABC$,if $r_1 = 2r_2 = 3r_3$,then the ratio $a : b$ is:

In a $\Delta ABC$,if ${c^2} + {a^2} - {b^2} = ac$,then $\angle B = $

In a $\triangle ABC$,$\frac{2(r_1+r_3)}{ac(1+\cos B)} = $

Assertion $(A)$: In $\triangle ABC$,if $r=6, r_2=36, R=15$,then $c^2+a^2=b^2$.
Reason $(R)$: In $\triangle ABC$,if $r:R:r_2=1:2.5:6$,then $B=90^{\circ}$.
The correct option among the following is:

With usual notations in a triangle $ABC$,if $r_1 = 2r_2 = 2r_3$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module