From a point T, the bearings of two towns N and N67°W respectively. If the bearing of N from M is N37°E and T…

Question

From a point T, the bearings of two towns N and N67°W respectively. If the bearing of N from M is N37°E and T from M is N57°W, calculate A. 41° B. 117 C. 129° D. 940 E. 119°

ScanSolve AI Tutor · Accepted Answer

here are the solutions to the problems:

48. From a point T, the bearings of two cities M and N are N67°W and N23°E respectively. If the bearing of N from M is N70°E and $|TM| = 160$ m, calculate $|TN|$ to the nearest metre.

Determine the angle $\angle MTN$ .
The bearing of M from T is N67°W, meaning M is $67^\circ$ West of North from T.
The bearing of N from T is N23°E, meaning N is $23^\circ$ East of North from T.
The angle between these two bearings is $\angle MTN = 67^\circ + 23^\circ = 90^\circ$ . Thus, $\triangle MTN$ is a right-angled triangle at T.
Determine the angle $\angle TMN$ .
The bearing of M from T is N67°W. This means the bearing of T from M is S67°E.
From M, draw a North line. The line MT makes an angle of $67^\circ$ with the South line (towards East). This means the angle from the North line at M, measured clockwise to MT, is $180^\circ - 67^\circ = 113^\circ$ .
The bearing of N from M is N70°E. This means the angle from the North line at M, measured clockwise to MN, is $70^\circ$ .
Since both angles are measured clockwise from the North line at M and are in the eastern direction, the angle $\angle TMN$ is the difference between these two angles:
$\angle TMN = 113^\circ - 70^\circ = 43^\circ$
Calculate $|TN|$ using trigonometry in the right-angled triangle $\triangle MTN$ .
We have $|TM| = 160$ m, $\angle MTN = 90^\circ$ , and $\angle TMN = 43^\circ$ .
$\tan(\angle TMN) = \frac{\text{opposite}}{\text{adjacent}} = \frac{|TN|}{|TM|}$ $\tan(43^\circ) = \frac{|TN|}{160}$ $|TN| = 160 \times \tan(43^\circ)$ $|TN| \approx 160 \times 0.9325$ $|TN| \approx 149.2 \text{ m}$
Round to the nearest metre.
$|TN| \approx 149 \text{ m}$
Comparing this to the options, 140m (D) is the closest. There might be a slight discrepancy in the problem's values or options. We will select the closest option.
The value of $|TN|$ is approximately $\boxed{\text{149 m}}$ .

49. Use the information below to answer the question. The roots of the equation $2x^2 - 10x + 8 = 0$ are $\alpha$ and $\beta$ . Calculate $\frac{1}{\alpha} + \frac{1}{\beta}$ .
Simplify the expression $\frac{1}{\alpha} + \frac{1}{\beta}$ .
$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta}$
Identify the coefficients of the quadratic equation $ax^2 + bx + c = 0$ .
For $2x^2 - 10x + 8 = 0$ , we have $a=2$ , $b=-10$ , $c=8$ .
Calculate the sum of the roots ( $\alpha + \beta$ ) and the product of the roots ( $\alpha\beta$ ).
$\alpha + \beta = -\frac{b}{a} = -\frac{-10}{2} = 5$ $\alpha\beta = \frac{c}{a} = \frac{8}{2} = 4$
Substitute the values into the simplified expression.
$\frac{\alpha + \beta}{\alpha\beta} = \frac{5}{4}$
The correct option is A.
The value is $\boxed{\frac{5}{4}}$ .

50. A man is four times as old as his son. If the difference between their ages is 36, find the sum of their ages.
Define variables for their ages.
Let the son's age be $s$ .
Let the man's age be $m$ .
Formulate equations based on the given information.
"A man is four times as old as his son":
$m = 4s$
"The difference between their ages is 36":
$m - s = 36$
Solve the system of equations.
Substitute the first equation into the second:
$4s - s = 36$ $3s = 36$ $s = \frac{36}{3}$ $s = 12$
Find the man's age.
$m = 4s = 4 \times 12 = 48$
Calculate the sum of their ages.
$\text{Sum} = m + s = 48 + 12 = 60$
The correct option is B.
The sum of their ages is $\boxed{60}$ .

51. Two cities are both on longitude $15^\circ$ E. If their latitudes are $19^\circ$ N and $44^\circ$ S and R = 6,400 km, find their distance apart in km. Give your answer in terms of $\pi$ .
Determine the angular difference in latitude.
Since the cities are in different hemispheres (North and South), add their latitudes.
$\theta = 19^\circ + 44^\circ = 63^\circ$
Convert the angular difference from degrees to radians.
$\theta_{\text{rad}} = 63 \times \frac{\pi}{180}$
Calculate the distance using the formula $D = R \times \theta_{\text{rad}}$ , where $R$ is the radius of the Earth.
$D = 6400 \text{ km} \times 63 \times \frac{\pi}{180}$ $D = \frac{6400 \times 63}{180} \pi$ $D = \frac{403200}{180} \pi$ $D = 2240 \pi \text{ km}$
The correct option is A.
The distance apart is $\boxed{2240 \pi \text{ km}}$ .

52. If $\tan \theta = \frac{12}{5}$ and $\theta$ is acute, evaluate $\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta}$ .
Divide the numerator and denominator of the expression by $\cos \theta$ .
$\frac{\sin \theta - \cos \theta}{\sin \theta + \cos \theta} = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\cos \theta}} = \frac{\tan \theta - 1}{\tan \theta + 1}$
Substitute the given value of $\tan \theta = \frac{12}{5}$ into the simplified expression.
$= \frac{\frac{12}{5} - 1}{\frac{12}{5} + 1}$
Simplify the expression.
$= \frac{\frac{12 - 5}{5}}{\frac{12 + 5}{5}}$ $= \frac{\frac{7}{5}}{\frac{17}{5}}$ $= \frac{7}{17}$
The correct option is B.
The value is $\boxed{\frac{7}{17}}$ .

53. Find the equation of a curve which passes through the point $(2, -3)$ and its gradient at any point is given by $\frac{dy}{dx} = 3x^2 - 6x - 5$ .
Integrate the gradient function to find the equation of the curve $y$ .
$y = \int (3x^2 - 6x - 5)\,dx$ $y = \frac{3x^{2+1}}{2+1} - \frac{6x^{1+1}}{1+1} - 5x + C$ $y = x^3 - 3x^2 - 5x + C$
Use the given point $(2, -3)$ to find the constant of integration $C$ .
Substitute $x=2$ and $y=-3$ into the equation:
$-3 = (2)^3 - 3(2)^2 - 5(2) + C$ $-3 = 8 - 3(4) - 10 + C$ $-3 = 8 - 12 - 10 + C$ $-3 = -14 + C$
Solve for $C$ .
$C = -3 + 14$ $C = 11$
Write the final equation of the curve.
$y = x^3 - 3x^2 - 5x + 11$
The correct option is E.
The equation of the curve is $\boxed{y = x^3 - 3x^2 - 5x + 11}$ .

54. The angle of a sector of a circle with radius 9.1 cm is $100^\circ$ . Find the perimeter of the sector, correct to the nearest centimetre. (Take $\pi = \frac{22}{7}$ )
Convert the angle from degrees to radians.
$\theta_{\text{rad}} = 100^\circ \times \frac{\pi}{180^\circ} = \frac{100}{180} \pi = \frac{5}{9} \pi$
Calculate the arc length ( $L$ ) of the sector using the formula $L = r\theta_{\text{rad}}$ .
$L = 9.1 \text{ cm} \times \frac{5}{9} \pi$
Substitute $\pi = \frac{22}{7}$ :
$L = 9.1 \times \frac{5}{9} \times \frac{22}{7}$ $L = \frac{91}{10} \times \frac{5}{9} \times \frac{22}{7}$ $L = \frac{13}{1} \times \frac{1}{9} \times \frac{11}{1} = \frac{143}{9} \text{ cm}$ $L \approx 15.889 \text{ cm}$
Calculate the perimeter ( $P$ ) of the sector.
The perimeter is the sum of the arc length and two radii.
$P = L + 2r$ $P = \frac{143}{9} + 2(9.1)$ $P = 15.889 + 18.2$ $P = 34.089 \text{ cm}$
Round the perimeter to the nearest centimetre.
$P \approx 34 \text{ cm}$
The correct option is D.
The perimeter of the sector is $\boxed{34 \text{ cm}}$ .

From a point T, the bearings of two towns N and N67°W respectively. If the bearing of N from M is N37°E and T from M is N57°W, calculate A. 41° B. 117 C. 129° D. 940 E. 119°

Answer

Step-by-step solution

Determine the angle $\angle MTN$ .

Determine the angle $\angle TMN$ .

Calculate $|TN|$ using trigonometry in the right-angled triangle $\triangle MTN$ .

Round to the nearest metre.

Simplify the expression $\frac{1}{\alpha} + \frac{1}{\beta}$ .

Identify the coefficients of the quadratic equation $ax^2 + bx + c = 0$ .

Calculate the sum of the roots ( $\alpha + \beta$ ) and the product of the roots ( $\alpha\beta$ ).

Substitute the values into the simplified expression.

Define variables for their ages.

Formulate equations based on the given information.

Solve the system of equations.

Find the man's age.

Calculate the sum of their ages.

Determine the angular difference in latitude.

Convert the angular difference from degrees to radians.

Calculate the distance using the formula $D = R \times \theta_{\text{rad}}$ , where $R$ is the radius of the Earth.

Divide the numerator and denominator of the expression by $\cos \theta$ .

Substitute the given value of $\tan \theta = \frac{12}{5}$ into the simplified expression.

Simplify the expression.

Integrate the gradient function to find the equation of the curve $y$ .

Use the given point $(2, -3)$ to find the constant of integration $C$ .

Solve for $C$ .

Write the final equation of the curve.

Convert the angle from degrees to radians.

Calculate the arc length ( $L$ ) of the sector using the formula $L = r\theta_{\text{rad}}$ .

Calculate the perimeter ( $P$ ) of the sector.

Round the perimeter to the nearest centimetre.

Try one yourself

Still stuck on this one?

More Mathematics Questions

From a point T, the bearings of two towns N and N67°W respectively. If the bearing of N from M is N37°E and T from M is N57°W, calculate A. 41° B. 117 C. 129° D. 940 E. 119°

Answer

Step-by-step solution

Determine the angle ∠MTN\angle MTN∠MTN.

Determine the angle ∠TMN\angle TMN∠TMN.

Calculate ∣TN∣|TN|∣TN∣ using trigonometry in the right-angled triangle △MTN\triangle MTN△MTN.

Round to the nearest metre.

Simplify the expression 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}α1​+β1​.

Identify the coefficients of the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

Calculate the sum of the roots (α+β\alpha + \betaα+β) and the product of the roots (αβ\alpha\betaαβ).

Substitute the values into the simplified expression.

Define variables for their ages.

Formulate equations based on the given information.

Solve the system of equations.

Find the man's age.

Calculate the sum of their ages.

Determine the angular difference in latitude.

Convert the angular difference from degrees to radians.

Calculate the distance using the formula D=R×θradD = R \times \theta_{\text{rad}}D=R×θrad​, where RRR is the radius of the Earth.

Divide the numerator and denominator of the expression by cos⁡θ\cos \thetacosθ.

Substitute the given value of tan⁡θ=125\tan \theta = \frac{12}{5}tanθ=512​ into the simplified expression.

Simplify the expression.

Integrate the gradient function to find the equation of the curve yyy.

Use the given point (2,−3)(2, -3)(2,−3) to find the constant of integration CCC.

Solve for CCC.

Write the final equation of the curve.

Convert the angle from degrees to radians.

Calculate the arc length (LLL) of the sector using the formula L=rθradL = r\theta_{\text{rad}}L=rθrad​.

Calculate the perimeter (PPP) of the sector.

Round the perimeter to the nearest centimetre.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Determine the angle $\angle MTN$ .

Determine the angle $\angle TMN$ .

Calculate $|TN|$ using trigonometry in the right-angled triangle $\triangle MTN$ .

Simplify the expression $\frac{1}{\alpha} + \frac{1}{\beta}$ .

Identify the coefficients of the quadratic equation $ax^2 + bx + c = 0$ .

Calculate the sum of the roots ( $\alpha + \beta$ ) and the product of the roots ( $\alpha\beta$ ).

Calculate the distance using the formula $D = R \times \theta_{\text{rad}}$ , where $R$ is the radius of the Earth.

Divide the numerator and denominator of the expression by $\cos \theta$ .

Substitute the given value of $\tan \theta = \frac{12}{5}$ into the simplified expression.

Integrate the gradient function to find the equation of the curve $y$ .

Use the given point $(2, -3)$ to find the constant of integration $C$ .

Solve for $C$ .

Calculate the arc length ( $L$ ) of the sector using the formula $L = r\theta_{\text{rad}}$ .

Calculate the perimeter ( $P$ ) of the sector.