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1 year ago

In Triangle 2, b=10 in, a=14 in, and theta=65 degrees solve for c, a, and B.

Mathematics

1 answer:

ziro4ka [17]1 year ago

6 0

The length of c is c = 13.3, and the angle measures are A = 17.7 and B = 97.3

<h3>How to determine the missing sides and angles of the triangle?</h3>

From the question, we have the given parameters to be:

b = 10 inches

a = 14 inches

<C = 65 degrees

To calculate the length of c, we use the following equation of law of cosine

c^2 = a^2 + b^2 - 2 * a * b * cos(C)

Substitute the known values in the above equation

c^2 = 14^2 + 10^2 - 2 * 10 * 14 * cos(65 degrees)

Evaluate cos(65 degrees)

c^2 = 14^2 + 10^2 - 2 * 10 * 14 * 0.4226

Evaluate the exponent

c^2 = 196 + 100 - 2 * 10 * 14 * 0.4226

Evaluate the product

c^2 = 196 + 100 - 118.328

Evaluate the sum and the difference

c^2 = 177.672

Take the square root of both sides

c = 13.3

To calculate the angle measure of A, we use the following equation of law of sine

a/sin(A) = c/sin(C)

This gives

14/sin(A) = 13.3/sin(65)

This gives

14/sin(A) = 14.7

Rewrite as:

sin(A) = 14/14.7

Evaluate the quotient

sin(A) = 0.9524

Take the arc sin of both sides

A = 17.7

Lastly, we have

B = 180 - A - C

This gives

B = 180 - 17.7 - 65

Evaluate

B = 97.3

Hence, the length of c is c = 13.3, and the angle measures are A = 17.7 and B = 97.3

Read more about law of cosines and sines at:

brainly.com/question/4372174

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These homogeneous recurrence relations of degree 2 have one of two solutions. Problems a, b, c, e, g have one solution; problems d and f have a slightly different solution. The solution method is similar, up to a point.

If there is a solution of the form $a[n]=r^n$ , then it will satisfy ...

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Rearranging and dividing by $r^{n-2}$ , we get the quadratic ...

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$r=\dfrac{c_1\pm\sqrt{c_1^2+4c_2}}{2}$

We can call these values of r by the names r₁ and r₂.

Then, for some coefficients p and q, the solution to the recurrence relation is ...

$a[n]=pr_1^n+qr_2^n$

We can find p and q by solving the initial condition equations:

$\left[\begin{array}{cc}1&1\\r_1&r_2\end{array}\right] \left[\begin{array}{c}p\\q\end{array}\right] =\left[\begin{array}{c}a[0]\\a[1]\end{array}\right]$

These have the solution ...

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And for problem (f), we get ...

$c_1=-6,\ c_2=-9,\ a[0]=3,\ a[1]=-3\\\\r=\dfrac{-6+\sqrt{6^2+4(-9)}}{2}=-3\\\\p=3,\ q=\dfrac{-3-3(-3)}{-3}=-2\\\\a[n]=(3-2n)(-3)^n$

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<em>Comment on problem g</em>

Yes, the bases of the exponential terms are conjugate irrational numbers. When the terms are evaluated, they do resolve to rational numbers.

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