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hoa [83]
2 years ago
10

A triangle with sides 11m , 13m and 18m is a right triangle. ​ A True B False

Mathematics
1 answer:
Julli [10]2 years ago
5 0

\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

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Neporo4naja [7]

Answer:

2.

sinα=√3/(2√7)

cosα=5/(2√7)

tanα=5/√3

cscα=(2√7)/√3

secα=(2√7)/5

*note* you can simplify the above further (by rationalizing them), but I think it's probably fine to leave them as is *

3.) g15.9

4.)25.4

5.) 70.8

6.) 36.4

Step-by-step explanation:

Cot is the inverse of toa so cot= adj/opp

This means the traingle's adjacent side is √3 and its opposite is 5

with this information let's figure out the hypotonouse

√3²+5²=c²

3+25=c²

28=c²

√28=c

√28=2√7

which means the triangle's

opposite= √3

hypotonous= 2√7

Adjacent= 5

With all of this we can just plug in the numbers to find the missing information (where α=angle or theta)

sinα=√3/(2√7)

cosα=5/(2√7)

tanα=5/√3

cscα=(2√7)/√3

secα=(2√7)/5

For this one we have the adjacent and need the opposite

we will use TOA

Tan(25)=x/34

34tan(25)=x

x=15.9

4.) For this one we have the adjacent but need the hypotonouse

we will use CAH

cos(48)=17/x

17/cos(48)=x

x=25.4

5.) for number 5 we have the oppsite and hypotonouse and so we'll use SOH

sin(α)=17/18

α=70.8

6.) For this one we have the opposite and adjacent and so we'll use TOA

Tan(α)=(31/42)

α=36.4

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