The angle of elevation of the top of a tower from a point A due north of it and from a point B at a distance of 9 units due west of A is <math><mrow><mi>c</mi><mi>o</mi><msup><mrow><mi>s</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo>(</mo><mrow><mfrac><mrow><mn>3</mn></mrow><mrow><mroot><mrow><mn>1</mn><mn>3</mn></mrow><mrow></mrow></mroot></mrow></mfrac></mrow><mo>)</mo></mrow><mo>.</mo></mrow></math> If the distance of the point B from the tower is 15 units, then cot is equal to:

100cot2α=152−92=144OT15=23cotα=65OT=10ΔAOT, OA=OTcotα=10cotα

The angle of elevation of the top of a tower from a point A due north of it and from a point B at a distance of 9 units due west of A is&nbsp;<math><mrow><mi>c</mi><mi>o</mi><msup><mrow><mi>s</mi></mrow><mrow><mo>−</mo><mn>1</mn></mrow></msup><mrow><mo>(</mo><mrow><mfrac><mrow><mn>3</mn></mrow><mrow><mroot><mrow><mn>1</mn><mn>3</mn></mrow><mrow></mrow></mroot></mrow></mfrac></mrow><mo>)</mo></mrow><mo>.</mo></mrow></math>&nbsp;If the distance of the point B from the tower is 15 units, then cot is equal to:

Ask & Answer Question

Option 1 -

$\frac{6}{5}$

Option 2 -

$\frac{9}{5}$

Option 3 -

$\frac{4}{3}$

Option 4 -

$\frac{7}{3}$

2 Views | Posted 2 months ago

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 1

Detailed Solution:

$100 c o t^{2} α = 1 5^{2} - 9^{2} = 144$

$\frac{O T}{15} = \frac{2}{3}$

$c o t α = \frac{6}{5} O T = 10$

$Δ A O T, O A = O T c o t α = 10 c o t α$

0 Upvotes 0 Downvotes

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to