Let P be an interior point of a triangle ABC. | vedclass

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

Question

(C) Let $\angle PAB = 	heta$ and $\angle PAC = \phi$. Since $Q$ is the reflection of $P$ in $AB$,we have $AQ = AP$ and $\angle QAB = \angle PAB = 	h

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) Let $\angle PAB = 	heta$ and $\angle PAC = \phi$. Since $Q$ is the reflection of $P$ in $AB$,we have $AQ = AP$ and $\angle QAB = \angle PAB = 	heta$. Since $R$ is the reflection of $P$ in $AC$,we have $AR = AP$ and $\angle RAC = \angle PAC = \phi$. Given that $Q, A, R$ are collinear,the angle $\angle QAR = 180^{\circ}$. From the figure,$\angle QAR = \angle QAB + \angle BAC + \angle RAC = 	heta + (	heta + \phi) + \phi = 2(	heta + \phi) = 180^{\circ}$. Thus,$	heta + \phi = 90^{\circ}$. Since $\angle BAC = 	heta + \phi$,we have $\angle BAC = 90^{\circ}$.

Let $P$ be an interior point of a $\triangle ABC$. Let $Q$ and $R$ be the reflections of $P$ in $AB$ and $AC$,respectively. If $Q, A, R$ are collinear,then $\angle A$ equals (in $^{\circ}$)

Similar Questions

Find the equation of the right bisector of the line segment joining the points $(3, 4)$ and $(-1, 2).$

The equation of the line bisecting perpendicularly the segment joining the points $(-4, 6)$ and $(8, 8)$ is

$A$ person standing at the junction (crossing) of $2$ straight paths represented by the equations $2x - 3y + 4 = 0$ and $3x + 4y - 5 = 0$,wants to reach the path whose equation is $6x - 7y + 8 = 0$ in the least time. The equation of the path he should follow is:

The image of the line $x+y-2=0$ in the $y$-axis is

The area (in square units) of the quadrilateral formed by the point of intersection of the lines $x+y-1=0$ and $x-y+1=0$,the point $P(1,1)$,and the feet of the perpendiculars from this point onto the lines is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module