Problem 42: The diagram shows a triangle LMN inscribed in a circle with center O. LN is the diameter. If LNM…

Question

Problem 42: The diagram shows a triangle LMN inscribed in a circle with center O. LN is the diameter. If LNM = (x+15)^ and NLM = (x−5)^, find x. Problem 43: Given vectors a = (3, 2) and b = (−1, 4), f

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Here are the solutions to the problems.

:
The diagram shows a triangle LMN inscribed in a circle with center O. Since LN passes through the center O, LN is the diameter of the circle. An angle inscribed in a semicircle is a right angle, so $\angle LMN = 90^\circ$ .

Set up the equation for the sum of angles in $\triangle LMN$ .
The sum of angles in a triangle is $180^\circ$ .
$\angle LNM + \angle NLM + \angle LMN = 180^\circ$
Substitute the given values: $\angle LNM = (x+15)^\circ$ , $\angle NLM = (x-5)^\circ$ , and $\angle LMN = 90^\circ$ .
$(x+15) + (x-5) + 90 = 180$
Solve for $x$ .
$2x + 10 + 90 = 180$ $2x + 100 = 180$ $2x = 180 - 100$ $2x = 80$ $x = \frac{80}{2}$ $x = 40$
The value of $x$ is 40.

The correct option is B.
$\boxed{\text{B. } 40^\circ}$
:
Given vectors $\mathbf{a} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ . We need to find $\frac{1}{2}\mathbf{a} - \mathbf{b}$ .
Calculate $\frac{1}{2}\mathbf{a}$ .
$\frac{1}{2}\mathbf{a} = \frac{1}{2} \begin{pmatrix} 3 \\ 2 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} \\ \frac{2}{2} \end{pmatrix} = \begin{pmatrix} 1.5 \\ 1 \end{pmatrix}$
Calculate $\frac{1}{2}\mathbf{a} - \mathbf{b}$ .
$\frac{1}{2}\mathbf{a} - \mathbf{b} = \begin{pmatrix} 1.5 \\ 1 \end{pmatrix} - \begin{pmatrix} -1 \\ 4 \end{pmatrix}$ $= \begin{pmatrix} 1.5 - (-1) \\ 1 - 4 \end{pmatrix}$ $= \begin{pmatrix} 1.5 + 1 \\ -3 \end{pmatrix}$ $= \begin{pmatrix} 2.5 \\ -3 \end{pmatrix}$
The correct option is A.
$\boxed{\text{A. } \begin{pmatrix} 2.5 \\ -3 \end{pmatrix}}$
:
Calculate the magnitude of the vector $2\mathbf{i} - 3\mathbf{j}$ .
Identify the components of the vector.
A vector $\mathbf{v} = x\mathbf{i} + y\mathbf{j}$ has components $x$ and $y$ .
For the vector $2\mathbf{i} - 3\mathbf{j}$ , we have $x = 2$ and $y = -3$ .
Calculate the magnitude using the formula $|\mathbf{v}| = \sqrt{x^2 + y^2}$ .
$|2\mathbf{i} - 3\mathbf{j}| = \sqrt{(2)^2 + (-3)^2}$ $= \sqrt{4 + 9}$ $= \sqrt{13}$
The magnitude of the vector $2\mathbf{i} - 3\mathbf{j}$ is $\sqrt{13}$ .
$\boxed{\sqrt{13}}$

Problem 42: The diagram shows a triangle LMN inscribed in a circle with center O. LN is the diameter. If LNM = (x+15)^ and NLM = (x−5)^, find x. Problem 43: Given vectors a = (3, 2) and b = (-1, 4), f

Answer

Step-by-step solution

Set up the equation for the sum of angles in $\triangle LMN$ .

Solve for $x$ .

Calculate $\frac{1}{2}\mathbf{a}$ .

Calculate $\frac{1}{2}\mathbf{a} - \mathbf{b}$ .

Identify the components of the vector.

Calculate the magnitude using the formula $|\mathbf{v}| = \sqrt{x^2 + y^2}$ .

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