Question

1. Bob tried to answer the following question by finding the missing angle and rounding the answer to the nearest degree.

Answer 1

Answer:

Part 1)

Bob's mistake was to have used the cosine instead of the sine

The measure of the missing angle is $53.13\°$

Part 2) The surface area of the pyramid is $288\ cm^{2}$

Step-by-step explanation:

Part 1)

Let

x----> the missing angle

we know that

In the right triangle o the figure

The sine of angle x is equal to divide the opposite side angle x to the hypotenuse of the right triangle

$sin(x)=\frac{16}{20}$

$x=arcsin(\frac{16}{20})=53.13\°$

Bob's mistake was to have used the cosine instead of the sine

Part 2) we know that

The surface area of the square pyramid is equal to the area of the square base plus the area of its four lateral triangular faces

so

$SA=b^{2}+4[\frac{1}{2}(b)(h)]$

where

b is the length side of the square

h is the height of the triangular lateral face

In this problem

$h=b/2$ -------> by an 45° angle

so

$b=2h$

$sin(45\°)=\frac{h}{6\sqrt{2}}$

$h=6\sqrt{2}(sin(45\°))=6\ cm$

Find the value of b

$b=2(6)=12\ cm$

Find the surface area

$SA=12^{2}+4[\frac{1}{2}(12)(6)]=288\ cm^{2}$