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2 years ago

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

Mathematics

1 answer:

hjlf2 years ago

5 0

Answer: 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.

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Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

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a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

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