# diagram 3D

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• December 15th 2011, 08:42 PM
ahmedb
diagram 3D
question: while standing at the left corner of the schoolyard in front of here school, Suzie estimates that the fronth face is 8.9m wide and 4.7 high. From her position. Suzie is 12.0m from the base of the right exterior wall. She determines that the let and the right exterior walls appear to be 39degrees apart. from her position, what is the angle of elevation, to the nearest degree, to the top of the left exterior wall?
can someone create the 3D diagram and show me how to solve what?
• December 16th 2011, 07:40 PM
Soroban
Re: diagram 3D
Hello, ahmedb!

The description is rather fuzzy; they should have supplied a diagram.
I hope I've interpreted it correctly/

Quote:

Qhile standing at the left corner of the schoolyard in front of her school,
Suzie estimates that the fronth face is 8.9m wide and 4.7 high.
From her position, Suzie is 12.0m from the base of the right exterior wall.
She determines that the left and the right exterior walls appear to be 39 degrees apart.
From her position, what is the angle of elevation, to the nearest degree,
. . to the top of the left exterior wall?

Code:

                D                  C                 o - - - - - - - - - o               *|                  |               * |4.7                |             *  |                  |             *  |      8.9        |           *    o - - - - - - - - - o           *  * A            *    B         *  *          *         * * 39d  * 12.0       **  *     S o
Rectangle $ABCD$ is the front face of the school.
. . $AB = 8.9,\;AD = 4.7$

Suzie is at point $S.\;\;\angle ASB = 39^o,\;SB = 12.0$

We want $\angle DSA.$

In $\Delta SAB$, we have: . $AB = 8.9,\;SB = 12.0,\;\angle S = 39^o$

Law of Sines: . $\frac{\sin A}{12} \,=\, \frac{\sin 39^o}{8.9} \quad\Rightarrow\quad \sin A \:=\:0.848521876$

Hence: . $A \:\approx\:58^o \quad\Rightarrow\quad B \:=\:180^o = 39^o - 58^o \:=\:83^o$

Law of sines: . $\frac{S\!A}{\sin83^o} \,=\,\frac{8.9}{\sin39^o} \quad\Rightarrow\quad S\!A \:=\:14.03682588$

Hence: . $S\!A \:\approx\:14$

In right triangle $D\!AS\!:\;\tan(\angle DSA) \,=\,\frac{4.7}{14} \:=\:0.335714285$

Therefore: . $\angle DSA \:=\:18.55763767 \;\approx\;18.6^o$