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2 years ago

HELP PLEASE HELP!!!!!

Mathematics

2 answers:

kotykmax [81]2 years ago

8 0

Answer:

(i): A triangle where all three sides are equal and the same.

(ii): A triangle where all three sides are unequal in length.

(iii): A triangle that has two sides of equal length.

(iv): A triangle where all three internal angles are acute(they measure less than 90 degrees.

(v): A triangle where all three internal angles are obtuse(measure above 90 degrees).

(vi): A triangle where one internal angle is a right angle(90 degrees).

Hope this helps

Step-by-step explanation:

bagirrra123 [75]2 years ago

3 0

(i): A triangle where all three sides are equal and the same.

(ii): A triangle where all three sides are unequal in length.

(iii): A triangle that has two sides of equal length.

(iv): A triangle where all three internal angles are acute(they measure less than 90 degrees.

(v): A triangle where all three internal angles are obtuse(measure above 90 degrees).

(vi): A triangle where one internal angle is a right angle(90 degrees).

Hope this helps

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Step-by-step explanation:

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We can use a ratio to solve

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

iren2701 [21]

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

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

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

$b=2h$

$sin(45\°)=\frac{h}{6\sqrt{2}}$

$h=6\sqrt{2}(sin(45\°))=6\ cm$

Find the value of b

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

Find the surface area

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