If in a triangle ABC, sin A, sin B, sin C are | vedclass

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

Question

(B) Given that $\sin A, \sin B, \sin C$ are in $A.P.$By the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the

Vedclass · Accepted Answer

the altitudes are in $H.P.$

Vedclass · Answer

(B) Given that $\sin A, \sin B, \sin C$ are in $A.P.$By the sine rule, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$, where $R$ is the circumradius.Thus, $a, b, c$ are in $A.P.$Let $p_1, p_2, p_3$ be the altitudes corresponding to sides $a, b, c$ respectively.The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.This implies $p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.Since $a, b, c$ are in $A.P.$, their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$Therefore, $\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$Hence, the altitudes $p_1, p_2, p_3$ are in $H.P.$

If in a triangle $ABC$, $\sin A, \sin B, \sin C$ are in $A.P.$, then

Similar Questions

In a triangle $ABC$,if $A = 2B$ and the sides opposite to the angles $A, B, C$ are $\alpha + 1, \alpha - 1$ and $\alpha$ respectively,then $\alpha =$

If the sides of a triangle are $3, 5, 7$,then

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

In $\triangle ABC, (r_2+r_3) \operatorname{cosec}^2 \frac{A}{2} =$

With usual notations, in $\triangle ABC$, if $\cos C = \frac{\sin A}{2 \sin B}$, then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module