If angles A, B and C are in A.P.,then frac{a+ | vedclass

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

Question

(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{\circ}$.Using

Vedclass · Accepted Answer

$2 \cos \frac{A-C}{2}$

Vedclass · Answer

(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$, we get $3B = 180^{\circ}$, so $B = 60^{\circ}$.Using the Sine Rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$, we have $a = k \sin A, b = k \sin B, c = k \sin C$.Therefore, $\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin B}$.Using the sum-to-product formula, $\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$.Since $A+C = 2B$, we have $\frac{A+C}{2} = B$.Thus, $\frac{a+c}{b} = \frac{2 \sin B \cos \left(\frac{A-C}{2}\right)}{\sin B} = 2 \cos \left(\frac{A-C}{2}\right)$.

If angles $A, B$ and $C$ are in $A$.$P$.,then $\frac{a+c}{b}$ is equal to

Similar Questions

Which of the following sets of data does not uniquely determine an acute-angled $\Delta ABC$ ($R$ = circum-radius)?

Let $ABC$ be a triangle with $\angle C=90^{\circ}$. Draw $CD$ perpendicular to $AB$. Choose points $M$ and $N$ on sides $AC$ and $BC$ respectively such that $DM$ is parallel to $BC$ and $DN$ is parallel to $AC$. If $DM=5$ and $DN=4$,then $AC$ and $BC$ are respectively equal to:

Let $ABC$ be a triangle such that $AB = BC$. Let $F$ be the mid-point of $AB$ and $X$ be a point on $BC$ such that $FX$ is perpendicular to $AB$. If $BX = 3XC$,then the ratio $BC / AC$ equals

The perimeter of $\Delta ABC$ is $6$ times the $A.M.$ of the sines of its angles. If $a = 1$,then $\angle A = $

In a $\triangle ABC$,if $\angle A = 3\angle B$,$CA = 9$,and $BC = 16$,then the length of $AB$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module