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Standard XII

Mathematics

Sum of n Terms

A vertical to...

Question

A vertical tower $P Q$ subtends the same angle $30^{o}$ at each of two places $A$ and $B$ , $60 m$ . apart on the ground, $A B$ subtends an angle $120^{o}$ at the foot of the tower. If $h$ is the height of the tower then $9 h^{2} + h + 1$ is equal to

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Solution

$A P = P B = h cot 30^{o} = \sqrt{3} h$
$cos 120^{o} = \frac{3 h^{2} + 3 h^{2} - 60^{2}}{2. 3 h^{2}} = - \frac{1}{2}$
$\Rightarrow 9 h^{2} = 60^{2} \Rightarrow h = 20 \Rightarrow 9 h^{2} + h + 1 = 3621$

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A vertical tower PQ subtends the same angle 30o at each of two places A and B, 60 m. apart on the ground, AB subtends an angle 120o at the foot of the tower. If h is the height of the tower then 9h2+h+1 is equal to

AP=PB=hcot30o=√3hcos120o=3h2+3h2−6022.3h2=−12⇒9h2=602⇒h=20⇒9h2+h+1=3621

A vertical tower $P Q$ subtends the same angle $30^{o}$ at each of two places $A$ and $B$ , $60 m$ . apart on the ground, $A B$ subtends an angle $120^{o}$ at the foot of the tower. If $h$ is the height of the tower then $9 h^{2} + h + 1$ is equal to

$A P = P B = h cot 30^{o} = \sqrt{3} h$
$cos 120^{o} = \frac{3 h^{2} + 3 h^{2} - 60^{2}}{2. 3 h^{2}} = - \frac{1}{2}$
$\Rightarrow 9 h^{2} = 60^{2} \Rightarrow h = 20 \Rightarrow 9 h^{2} + h + 1 = 3621$