Question

Question 21? Please?

Answer 1

The angle of elevation is how much Kayla has to angle her head upward to look directly at the plane. The angle immediately adjacent to that (the bottom interior angle of the triangle) is complementary to the angle of elevation. So when she first looks, this triangle is such that

$\tan(90^\circ-67^\circ40')=\dfrac a{880\,\rm m}$

where $a$ is the distance the plane is away from Kayla if she were standing 880 m above the ground. Solving this gives approximately

$a=(880\,\mathrm )\tan(90^\circ-67^\circ40')\approx361.5\,\mathrm m$

The next time Kayla looks up, she and the plane form a triangle such that

$\tan(90^\circ-24^\circ30')=\dfrac b{880\,\rm m}$

$b=(880\,\mathrm m)\tan(90^\circ-24^\circ30')\approx1931\,\mathrm m$

(a) This means that between the two times Kayla looks up, the plane traveled about 1569 m (compute the difference $b-a$).

(b) Assuming the plane moves at a constant speed $v$, after 25 s it will have traveled a distance of $(25\,\mathrm s)v$. So we have

$b-a=(25\,\mathrm s)v\implies v=\dfrac{1569\,\rm m}{25\,\rm s}=\dfrac{(1569\,\mathrm m)\left(\frac1{1000}\frac{\rm km}{\rm m}\right)}{(25\,\mathrm s)\left(\frac1{3600}\frac{\rm h}{\rm s}\right)}\approx226\dfrac{\rm km}{\rm h}$