A piece if wire 40cm is bent to form a right angled triangle whose hypotenuse is 17cm long find the other two sides of triangle

150mm

Step-by-step explanation:

Take the two adjecent sides of the triangle to be x and y and X is the angle opposite the side, x.

From the trigonometric identities for a right-angled triangle, it is known that

x=170sin X and;

y=170cos X

Thus the perimeter of the triangle is given by

170sin X + 170cos X + 170 = 400

170(sin X + cos X) = 400 - 170

170(sin X + cos X) = 230

sin X + cos X = 230/170 = 1.3529

Squaring both sides hence,

sin^2(X) + 2sin X.cos X + cos^2(X) = 1.8304

Recall that sin^2(X) + cos^2(X) = 1 and 2sin X.cos X = sin 2X.

Thus, the former expression gives;

1 + sin 2X = 1.8304

sin 2X = 0.8304

2X = arcsin (0.8304)

2X = 56.14° or (180 - 56.14)° = 56.14° or 123.86° for X existing within a triangle.

X = 56.14/2 or 123.86/2

X = 28.07° or 61.93°

Thus the two adjacent sides are ;

x = 170 sin (28.07) = 80mm

y = 170 cos (28.07) = 150mm

Using the value of 61.93° as X would give the same values but interchanged for x and y.

I hope you find this useful.

whole numbers.

step-by-step explanation:

we are given the average monthly rainfall (in millimeters) model r(m) for the years 1900-2009 as a function of the month number m.

as we can see m is taken for the number of months for years 1900-2009.

we know that number of months can't be a negative number or decimal number or a fraction.

so, the number of months could be only positive whole numbers.

therefore, whole numbers is more appropriate for the domain of r.

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