1. ## Trigonometry Problem 3

To calculate the height of a crane which is on top of a building, Denis measures the angle of elevation to the bottom and top of the crane. These were 62° and 68° respectively. If the building is 42 m high find, to 2 decimal places:
a) how far Denis is from the building
b) the height of the crane

2. Originally Posted by Joker37
To calculate the height of a crane which is on top of a building, Denis measures the angle of elevation to the bottom and top of the crane. These were 62° and 68° respectively. If the building is 42 m high find, to 2 decimal places:
a) how far Denis is from the building
b) the height of the crane
1. You are dealing with 2 right triangles (see attachment)

2. $\displaystyle \tan(62^\circ)=\dfrac{42}{x}$ . Solve for x.

3. The length of the crane is

$\displaystyle x \cdot \tan(68^\circ) - 42 = l_{crane}$

3. Hello, Joker37!

Did you make a sketch?

To calculate the height of a crane which is on top of a building,
Denis measures the angle of elevation to the bottom and top of the crane.
These were 62° and 68° respectively.
If the building is 42 m high find, to 2 decimal places:

a) how far Denis is from the building

b) the height of the crane.
Code:
    A o
|\
| \
y |  \
|   \
|    \
B o     \
| *    \
|   *   \
42 |     *6°\
|       * \
|     62° *\
C o - - - - - o D
x
The crane is $\displaystyle AB = y.$
The building is $\displaystyle BC = 42.$
Denis is at $\displaystyle D\!:\;CD = x$
$\displaystyle \angle BDC = 62^o,\;\angle ADC = 68^o$

(a) In right triangle $\displaystyle BCD\!:\;\;\tan 62^o \:=\:\frac{42}{x} \quad\Rightarrow\quad x \:=\:\frac{42}{\tan62^o} \;\approx\;22.33$ ft.

(b) In right triangle $\displaystyle ACD\!:\;\;\tan68^o \:=\:\frac{y+42}{x} \quad\Rightarrow\quad y \:=\:x\tan68^o - 42 \;\approx\;13.27$ ft.

Edit: Too slow again . . . *sigh*
.