Question

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

(see photo)
Here is his solution
cosx =16/20
x=cos-1 (16/20) = 36.8698976 = 37degree
Bob made a mistake in his work. Explain the mistake AND write the correct solution.

2) The following picture is a square pyramid where AE=6/2 cm and angle EAB =45degrees.
Find the surface area of the pyramid. (Make sure you include formulas and numbers in the formulas as part of your solution. Also include units in your final answer.)

Answer 1

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)`

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)=(16)/(20)`

`x=arcsin((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[(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\°)=(h)/(6√(2))`

`h=6√(2)(sin(45\°))=6\ cm`

Find the value of b

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

Find the surface area

`SA=12^(2)+4[(1)/(2)(12)(6)]=288\ cm^(2)`