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2 years ago

What are the measures of 1 and 2? 37.1 1 2 17.30 121.8°

Mathematics

2 answers:

Lady_Fox [76]2 years ago

8 0

Answer:

Option 2 : m∠1 = 67.4°, m∠2 = 104.5°

Step-by-step explanation:

Lets find the measure of angle 2 first:

180° - 17.3° - (180° - 121.8°) = Angle 2

↑

Angle of the straight line = 180°

The angle should be 180 - 121.8

Angle 2 = 104.5

At this point the only answer that works is the second option, but let's still work out Angle 1

Angle 1 :

Theory of opposite angles state that the opposite angles inside a triangle = the exterior angle

in this case:

Angle 1 + 37.1 = Angle 2

Since we know angle 2 = 104.5, lets solve

Lets say angle 1 = x

x + 37.1 = 104.5

x = 104.5 - 37.1

x = 67.4

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-Chetan K

Ugo [173]2 years ago

4 0

The measures of ∠ 1 and ∠ 2.

<h2><u>Solution</u>:-</h2>

${\boxed{\mathcal{\red{Angle\: 1 \:=\: 67.4° \:}}}}$ ✅

${\boxed{\mathcal{\red{Angle\: 2\:=\:104.5° \:}}}}$ ✅

<h2><u>Step</u>-<u>by</u>-<u>step</u> explanation:-</h2>

We know that,

$\sf\purple{Sum\:of\:angles\:on\:a\:straight\:line\:=\:180°}$

✒ ∠ x + 121.8° = 180°

✒ ∠ x = 180° - 121.8°

✒ ∠ x = 58.2°

Now,

$\sf\pink{Sum\:of\:angles\:of\:a\:triangle\:=\:180°}$

➪ ∠ 2 + 17.3° + 58.2° = 180°

➪ ∠ 2 + 75.5° = 180°

➪ ∠ 2 = 180° - 75.5°

➪ ∠ 2 = 104.5°

$\boxed{Therefore,\:the\:measure\:of\:∠\:2\:is\:104.5°.}$

Again,

$\sf\blue{Sum\:of\:angles\:on\:a\:straight\:line\:=\:180°}$

➵ ∠ 2 + ∠ y = 180°

➵ 104.5° + ∠ y = 180°

➵ ∠ y = 180° - 104.5°

➵ ∠ y = 75.5°

Finally, we have

$\sf\pink{Sum\:of\:angles\:of\:a\:triangle\:=\:180°}$

⇢ ∠ 1 + 75.5° + 37.1° = 180°

⇢ ∠ 1 + 112.6° = 180°

⇢ ∠ 1 = 180° - 112.6°

⇢ ∠ 1 = 67.4°

$\boxed{Therefore,\:the\:measure\:of\:∠\:1\:is\:67.4°.}$

$\bold{ \green{ \star{ \orange{Hope\:it\:helps.}}}}⋆$

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If you had 1052 toothpicks and were asked to group them in powers of 6, how many groups of each power of 6 would you have? Put t

sukhopar [10]

1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

The number 1052, written as a base 6 number is 4512

Given: 1052 toothpicks

To do: The objective is to group the toothpicks in powers of 6 and to write the number 1052 as a base 6 number

First we note that, $6^{0}=1,6^{1}=6,6^{2}=36,6^{3}=216,6^{4}=1296$

This implies that $6^{4}$ exceeds 1052 and thus the highest power of 6 that the toothpicks can be grouped into is 3.

Now, $6^{3}=216$ and $216\times 5=1080, 216\times 4=864$ . This implies that $216\times 5$ exceeds 1052 and thus there can be at most 4 groups of $6^{3}$ .

Then,

$1052-4\times6^{3}$

$1052-4\times216$

$1052-864$

$188$

So, after grouping the toothpicks into 4 groups of third power of 6, there are 188 toothpicks remaining.

Now, $6^{2}=36$ and $36\times 5=180, 36\times 6=216$ . This implies that $36\times 6$ exceeds 188 and thus there can be at most 5 groups of $6^{2}$ .

Then,

$188-5\times6^{2}$

$188-5\times36$

$188-180$

$8$

So, after grouping the remaining toothpicks into 5 groups of second power of 6, there are 8 toothpicks remaining.

Now, $6^{1}=6$ and $6\times 1=6, 6\times 2=12$ . This implies that $6\times 2$ exceeds 8 and thus there can be at most 1 group of $6^{1}$ .

Then,

$8-1\times6^{1}$

$8-1\times6$

$8-6$

$2$

So, after grouping the remaining toothpicks into 1 group of first power of 6, there are 2 toothpicks remaining.

Now, $6^{0}=1$ and $1\times 2=2$ . This implies that the remaining toothpicks can be exactly grouped into 2 groups of zeroth power of 6.

This concludes the grouping.

Thus, it was obtained that 1052 toothpicks can be grouped into 4 groups of third power of 6 ( $6^{3}$ ), 5 groups of second power of 6 ( $6^{2}$ ), 1 group of first power of 6 ( $6^{1}$ ) and 2 groups of zeroth power of 6 ( $6^{0}$ ).

Then,

$1052=4\times6^{3}+5\times6^{2}+1\times6^{1}+2\times6^{0}$