If two vertical poles of heights 20,m and 80, | vedclass

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

Question

(A) Let the two poles be $OA$ and $BC$ standing on a horizontal plane $OB$. Let $OA = 20\,m$ and $BC = 80\,m$. Let the distance between them be $OB =

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Let the two poles be $OA$ and $BC$ standing on a horizontal plane $OB$. Let $OA = 20\,m$ and $BC = 80\,m$. Let the distance between them be $OB = a$.We set the coordinates as $O(0, 0)$,$A(0, 20)$,$B(a, 0)$,and $C(a, 80)$.The line joining the top of the first pole $A(0, 20)$ to the foot of the second pole $B(a, 0)$ has the equation:$y - 0 = \frac{20 - 0}{0 - a}(x - a) \Rightarrow y = -\frac{20}{a}(x - a) \quad \dots(1)$The line joining the top of the second pole $C(a, 80)$ to the foot of the first pole $O(0, 0)$ has the equation:$y - 0 = \frac{80 - 0}{a - 0}(x - 0) \Rightarrow y = \frac{80}{a}x \quad \dots(2)$To find the intersection point $M$,we equate $(1)$ and $(2)$:$\frac{80}{a}x = -\frac{20}{a}(x - a)$$80x = -20x + 20a$$100x = 20a \Rightarrow x = \frac{a}{5}$Substituting $x = \frac{a}{5}$ into $(2)$:$y = \frac{80}{a} 	imes \frac{a}{5} = 16\,m$.Thus,the height of the point of intersection is $16\,m$.

If two vertical poles of heights $20\,m$ and $80\,m$ stand apart on a horizontal plane,then the height (in $m$) of the point of intersection of the lines joining the top of each pole to the foot of the other is:

Similar Questions

The line $lx + my + n = 0$ will be parallel to the $x$-axis,if

In the figure,$AHKF$,$FKDE$ and $HBCK$ are unit squares. $AD$ and $BF$ intersect at $X$. Then,the ratio of the areas of triangles $AXF$ and $ABF$ is

The ends of the base of an isosceles triangle are at $(2a, 0)$ and $(0, a)$. The equation of one side is $x = 2a$. The equation of the other side is

The equation of a straight line passing through the points $(-5, -6)$ and $(3, 10)$ is:

Two lines $L_1$ and $L_2$ passing through the point $P(1, 2)$ cut the line $x+y=4$ at a distance of $\frac{\sqrt{6}}{3}$ units from $P$. Then the angles made by $L_1$ and $L_2$ with the positive $X$-axis are

Vedclass Test Series

Exam Paper Generator

Online Exam Module