Two vertical poles A B=15 m and CD =10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD , then the height of P ( in m ) above the line AC is :

Question

Two vertical poles A B=15 m and CD =10 m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD , then the height of P ( in m ) above the line AC is : (A) 20 / 3 (B) 5 (C) 10 / 3 (D) 6

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/02ae1b9f8e65bc63d6328c97958083f2-.png" /> tan θ=(10/x)=(h/x2) ⇒ x2=(h x/10) tan φ=(15/x)=(h/x1) ⇒ x1=(h x/15) Now, x1+x2=x=(h x/15)+(h x/10) ⇒ 1=(h/10)+(h/15) ⇒ h=6

Q. Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points A and $C$ on the ground. If $P$ is the point of intersection of $BC$ and $A D,$ then the height of $P ($ in $m)$ above the line $A C$ is :

$20/3$

5

$10/3$

6

$tan θ = \frac{10}{x} = \frac{h}{x _{2}}$
$\Rightarrow x_{2} = \frac{h x}{10}$
$tan ϕ = \frac{15}{x} = \frac{h}{x _{1}}$
$\Rightarrow x_{1} = \frac{h x}{15}$
Now, $x_{1} + x_{2} = x = \frac{h x}{15} + \frac{h x}{10}$
$\Rightarrow 1 = \frac{h}{10} + \frac{h}{15}$
$\Rightarrow h = 6$

Q. Two vertical poles AB=15m and CD=10m are standing apart on a horizontal ground with points A and C on the ground. If P is the point of intersection of BC and AD, then the height of P( in m) above the line AC is :

20/3

5

10/3

6

tanθ=x10​=x2​h​ ⇒x2​=10hx​ tanϕ=x15​=x1​h​ ⇒x1​=15hx​ Now, x1​+x2​=x=15hx​+10hx​ ⇒1=10h​+15h​ ⇒h=6

Q. Two vertical poles $A B = 15 m$ and $C D = 10 m$ are standing apart on a horizontal ground with points A and $C$ on the ground. If $P$ is the point of intersection of $BC$ and $A D,$ then the height of $P ($ in $m)$ above the line $A C$ is :

$20/3$

$10/3$

$tan θ = \frac{10}{x} = \frac{h}{x _{2}}$
$\Rightarrow x_{2} = \frac{h x}{10}$
$tan ϕ = \frac{15}{x} = \frac{h}{x _{1}}$
$\Rightarrow x_{1} = \frac{h x}{15}$
Now, $x_{1} + x_{2} = x = \frac{h x}{15} + \frac{h x}{10}$
$\Rightarrow 1 = \frac{h}{10} + \frac{h}{15}$
$\Rightarrow h = 6$