Question

assignment

1. from the top of a building 21.0 m tall, the angle of elevation of the top of a taller building is 46°. the angle of depression of the base of the taller building is 51°. what is the height of the taller building?
2. find the length of ab.
3. find the length of ad. show the steps of your solution.

Explanation:

Step1: Solve for the distance between buildings in first problem

In the first - building problem, in right - triangle BDE, $\tan(51^{\circ})=\frac{BE}{DE}$, where $BE = 21$ m. So, $DE=\frac{21}{\tan(51^{\circ})}$.

Step2: Solve for the extra height of the taller building

In right - triangle ABC, $\tan(46^{\circ})=\frac{AC}{BC}$, and $BC = DE$. So, $AC = DE\times\tan(46^{\circ})=\frac{21\times\tan(46^{\circ})}{\tan(51^{\circ})}$.

Step3: Calculate the height of the taller building

The height of the taller building $h=21 + AC=21+\frac{21\times\tan(46^{\circ})}{\tan(51^{\circ})}$.
$h = 21+\frac{21\times1.03553}{1.23489}\approx21 + 17.5=38.5$ m.

Step4: Solve for the length of AB in the second problem

We can use the Pythagorean theorem. The vertical distance from the top - left point to the point on the same horizontal as A is $3$ m, and the horizontal distance from the right - most point to the point on the same vertical as A is $10$ m. The length of the hypotenuse of the right - triangle formed by the 6 - m slant and the vertical and horizontal distances related to it: Let's first find the vertical and horizontal components of the 6 - m slant. The vertical component of the 6 - m slant is $6\sin(40^{\circ})\approx6\times0.6428 = 3.8568$ m and the horizontal component is $6\cos(40^{\circ})\approx6\times0.7660 = 4.596$ m.
The total vertical distance from A to B is $3 + 3.8568=6.8568$ m and the total horizontal distance is $10 - 4.596 = 5.404$ m.
$AB=\sqrt{(6.8568)^{2}+(5.404)^{2}}\approx\sqrt{47.02 + 29.20}\approx\sqrt{76.22}\approx8.73$ m.

Step5: Solve for the length of AD in the third problem

First, in right - triangle ABC, $\tan(48^{\circ})=\frac{AB}{BC}$, with $BC = 55$ m. So, $AB = 55\times\tan(48^{\circ})\approx55\times1.1106 = 61.083$ m.
In right - triangle ACD, $\tan(55^{\circ})=\frac{AC}{CD}$, and we know that in right - triangle ABC, $AC = AB$.
In right - triangle ACD, $CD=\frac{AB}{\tan(55^{\circ})}=\frac{61.083}{1.4281}\approx42.77$ m.
Using the Pythagorean theorem in right - triangle ACD, $AD=\sqrt{AB^{2}+CD^{2}}=\sqrt{(61.083)^{2}+(42.77)^{2}}=\sqrt{3731.1+1830.2}=\sqrt{5561.3}\approx74.6$ m.

Answer:

1. The height of the taller building is approximately $38.5$ m.
2. The length of AB is approximately $8.73$ m.
3. The length of AD is approximately $74.6$ m.