If the median of Delta ABC through A is perpe | vedclass

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

Question

(C) Let $AD$ be the median through $A$ to the side $BC$. We are given that $AD \perp AB$,so $\angle DAB = 90^{\circ}$.Draw $CN \perp AB$ (produced). I

Vedclass · Accepted Answer

$	an A + 2	an B = 0$

Vedclass · Answer

(C) Let $AD$ be the median through $A$ to the side $BC$. We are given that $AD \perp AB$,so $\angle DAB = 90^{\circ}$.Draw $CN \perp AB$ (produced). In $\Delta BCN$,$AD \parallel CN$ because both are perpendicular to $AB$.Since $D$ is the midpoint of $BC$,by the intercept theorem,$A$ is the midpoint of $BN$,so $AB = AN$.In $\Delta ACN$,$\angle CAN = \alpha$. Then $\angle CAB = 180^{\circ} - \alpha$,so $	an A = 	an(180^{\circ} - \alpha) = -	an \alpha$.In $\Delta ACN$,$	an \alpha = \frac{CN}{AN}$.In $\Delta ABD$,$	an B = \frac{AD}{AB}$.Since $AD = \frac{1}{2}CN$ and $AB = AN$,we have $	an \alpha = \frac{2AD}{AB} = 2	an B$.Therefore,$	an A = -2	an B$,which implies $	an A + 2	an B = 0$.

If the median of $\Delta ABC$ through $A$ is perpendicular to $AB$,then

Similar Questions

With usual notations,in $\triangle ABC$,the lengths of two sides are $10 \text{ cm}$ and $9 \text{ cm}$ respectively. If angles $A, B, C$ are in $A$.$P$.,then the perimeter of $\triangle ABC$ is:

In a $\triangle ABC$,if $\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$,then $B$ is equal to

In $\triangle ABC$,find the value of $\sqrt{\frac{r r_2}{r_3 r_1}}$.

In $\Delta ABC,$ $\text{cosec } A(\sin B \cos C + \cos B \sin C) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module