A vertical pole fixed to the ground is divided in the ratio 1:9 by a mark on it with lower part shorter than the upper part. If the two parts subtend equal angles at a place on the ground, 15 m away from the base of the pole, what is the height of the pole?
60√5m
15√5m
15√3 m
60√3m
Explanation:
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that
$\angle$CAD =$\angle$BAD = θ
From "Angle Bisector Theorem", we have
$\dfrac{BD}{DC}=\dfrac{AB}{AC}$
⇒$\dfrac{1}{9}=\dfrac{15}{AC}$ [∵ BD : DC = 1 : 9 and AB = 15(given)]
=> AC = 15 × 9 m ...(eq: 1)
From the right $\triangle$ ABC,
CB=$\sqrt{AC^{2}-AB^{2}}$ (∵ Pythagorean theorem)
=$\sqrt{(15 \times 9)^{2}-15^{2}}$ (∵AC=15×9(eq:1) and AB=15 m(given))
=$\sqrt{15^{2} \times 9^{2}-15^{2}}$
=$\sqrt{15^{2} (9^{2}-1)}$=$\sqrt{15^{2} \times 80}$
=$\sqrt{15^{2} \times 16 \times 5}$ =15×4×√5
=60√5 m
Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that
$\angle$CAD =$\angle$BAD = θ
From "Angle Bisector Theorem", we have
$\dfrac{BD}{DC}=\dfrac{AB}{AC}$
⇒$\dfrac{1}{9}=\dfrac{15}{AC}$ [∵ BD : DC = 1 : 9 and AB = 15(given)]
=> AC = 15 × 9 m ...(eq: 1)
From the right $\triangle$ ABC,
CB=$\sqrt{AC^{2}-AB^{2}}$ (∵ Pythagorean theorem)
=$\sqrt{(15 \times 9)^{2}-15^{2}}$ (∵AC=15×9(eq:1) and AB=15 m(given))
=$\sqrt{15^{2} \times 9^{2}-15^{2}}$
=$\sqrt{15^{2} (9^{2}-1)}$=$\sqrt{15^{2} \times 80}$
=$\sqrt{15^{2} \times 16 \times 5}$ =15×4×√5
=60√5 m