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Misha Larkins [42]
3 years ago
5

Use a calculator to find the values of the inverse function in radians. sin-1 (0.65)

Mathematics
2 answers:
Bond [772]3 years ago
3 0
We are given the inverse function:
sin-1 (0.65)

We can use a scientific calculator to solve this. Make sure that the mode is in radians and not in degrees
So, the answer is
sin-1 (0.65) = 0.7076 which is in radians
Ad libitum [116K]3 years ago
3 0

Answer:

\sin^{-1}(0.65)=0.7075  

Step-by-step explanation:

Given : Inverse function \sin^{-1}(0.65)

To find : The values of the inverse function in radians?  

Solution :

Write the inverse function \sin^{-1}(0.65)  

Using calculator we find the value of inverse function in radians is

\sin^{-1}(0.65)=0.7075  

If the calculator calculate in degree the first find in degree and to convert into radian multiply its by \frac{\pi }{180}

In degree, \sin^{-1}(0.65)=40.5416  

In radian, \sin^{-1}(0.65)=40.5416\times \frac{\pi }{180}=0.7075

Therefore, The value of inverse function is \sin^{-1}(0.65)=0.7075  

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find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM​
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\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}

  • Given - <u>A </u><u>trapezium</u><u> </u><u>ABCD </u><u>with </u><u>non </u><u>parallel </u><u>sides </u><u>of </u><u>measure </u><u>1</u><u>5</u><u> </u><u>cm </u><u>each </u><u>!</u><u> </u><u>along </u><u>,</u><u> </u><u>the </u><u>parallel </u><u>sides </u><u>are </u><u>of </u><u>measure </u><u>1</u><u>3</u><u> </u><u>cm </u><u>and </u><u>2</u><u>5</u><u> </u><u>cm</u>

  • To find - <u>Area </u><u>of </u><u>trapezium</u>

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

<u>Construction</u><u> </u><u>-</u>

draw \: CE \: \parallel \: AD \:  \\ and \: CD \: \perp \: AE

Now , we can clearly see that AECD is a parallelogram !

\therefore AE = CD = 13 cm

Now ,

AB = AE + BE \\\implies \: BE =AB -  AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm

Now , In ∆ BCE ,

semi \: perimeter \: (s) =  \frac{15 + 15 + 12}{2}  \\  \\ \implies \: s =  \frac{42}{2}  = 21 \: cm

Now , by Heron's formula

area \: of \: \triangle \: BCE =  \sqrt{s(s - a)(s - b)(s - c)}  \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)}  \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21}  \: cm {}^{2}

Also ,

area \: of \: \triangle \:  =  \frac{1}{2}  \times base \times height \\  \\\implies 18 \sqrt{21} =  \: \frac{1}{\cancel2}  \times \cancel12  \times height \\  \\ \implies \: 18 \sqrt{21}  = 6 \times height \\  \\ \implies \: height =  \frac{\cancel{18} \sqrt{21} }{ \cancel 6}  \\  \\ \implies \: height = 3 \sqrt{21}  \: cm {}^{2}

<u>Since </u><u>we've </u><u>obtained </u><u>the </u><u>height </u><u>now </u><u>,</u><u> </u><u>we </u><u>can </u><u>easily </u><u>find </u><u>out </u><u>the </u><u>area </u><u>of </u><u>trapezium </u><u>!</u>

Area \: of \: trapezium =  \frac{1}{2}  \times(sum \: of \:parallel \: sides) \times height \\  \\ \implies \:  \frac{1}{2}  \times (25 + 13) \times 3 \sqrt{21}  \\  \\ \implies \:  \frac{1}{\cancel2}  \times \cancel{38 }\times 3 \sqrt{21}  \\  \\ \implies \: 19 \times 3 \sqrt{21}  \: cm {}^{2}  \\  \\ \implies \: 57 \sqrt{21}  \: cm {}^{2}

hope helpful :D

6 0
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Solve for x 5 – 57− 2 = 63 please help
umka2103 [35]
5x - 2x = 63 + 57
3x = 120
x = 40
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