The figure alongside is a square-based pyramid. Find: a) \(\angle ADM\) b) \(\angle ACD\).
Understand the Problem
The question is asking to find angle (\angle ADM) and (\angle ACD) in the given square-based pyramid. We are given the side length of the square base (20 cm) and the height of the pyramid (27 cm).
Answer
a. $\angle ADM \approx 69.66^\circ$ b. $\angle ACD = 45^\circ$
Answer for screen readers
a. $\angle ADM \approx 69.66^\circ$ b. $\angle ACD = 45^\circ$
Steps to Solve
- Find $DM$
Since $ABCD$ is a square with side length 20 cm, $M$ is the midpoint of $CD$. Therefore, $DM = \frac{20}{2} = 10$ cm.
- Find $\angle ADM$
We can use the tangent function in the right triangle $ADM$, where $AM = 27$ cm and $DM = 10$ cm. $$ \tan(\angle ADM) = \frac{AM}{DM} = \frac{27}{10} $$ $$ \angle ADM = \arctan\left(\frac{27}{10}\right) \approx 69.66^\circ $$
- Find $AC$
Since $ABCD$ is a square, we can use the Pythagorean theorem to find the length of the diagonal $AC$: $$ AC = \sqrt{AD^2 + DC^2} = \sqrt{20^2 + 20^2} = \sqrt{400 + 400} = \sqrt{800} = 20\sqrt{2} \text{ cm} $$
- Find $\angle ACD$
Since $ABCD$ is a square, all angles are $90^\circ$ and the diagonals bisect these angles. Therefore, $\angle ACD = \frac{90^\circ}{2} = 45^\circ$.
a. $\angle ADM \approx 69.66^\circ$ b. $\angle ACD = 45^\circ$
More Information
The angle $\angle ADM$ is the angle between the slant edge $AD$ and the line segment $DM$ on the base of the pyramid. The angle $\angle ACD$ is the angle formed by two adjacent vertices ($A$ and $C$) and the vertex $D$ of the square base.
Tips
A common mistake is to incorrectly identify the sides of the right triangle when finding $\angle ADM$. Ensure that you correctly identify the opposite and adjacent sides relative to the angle you are trying to find. Another common mistake is to try to use trigonometry to find $\angle ACD$, when simply understanding properties of a square will give you the answer directly.
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