Question

A student standing exactly midway between two buildings, observes the top of the two buildings at an angle of elevation of 22.5 and 67.5. What is the ratio of the height of the taller building to the height of the shorter buildings? (given that tan 22.5= v2-1)

Answer 1

Answer:

$\frac{1}{(\sqrt{2}-1 )^{2} }$

Step-by-step explanation

given

tan 22.5=$\sqrt{2}-1$

tan 67.5=cot (90-67.5)=$\frac{1}{tan 22.5}$=$\frac{1}{\sqrt{2} - 1 }$

let the distance between the buildings is x,hencw the boy is $\frac{x}{2}$ away from each building

height of taller building =h1

height of shorter nuilding=h2

since tan A =$\frac{perpendicular}{base}$

hence from the fig.

tan 22.5=$\frac{h2}{\frac{x}{2} }$

h2=$\sqrt{2}-1 \times \frac{x}{2}$ .........(1)

now, tan 67.5=$\frac{h1}{\frac{x}{2} }$

h1=$tan 67.5\times \frac{x}{2}$.......(2)

from eq 1 and eq 2,

$\frac{h1}{h2}=\frac{tan67.5 \times\frac{x}{2} }{tan22.5 \times\frac{x}{2} }$

$\frac{h1}{h2}=\frac{1}{(\sqrt{2}-1 )\times (\sqrt{2}-1 )}$

hence ratio of height of taller building to shorter building=$\frac{1}{(\sqrt{2}-1 )^{2} }$