Which one is it A, B, C, or D?

In this figure , m<1=(2x)° and m<2 = (6x-6)° what is m<1

mojhsa [17]

Answer:

where is the figure?

Step-by-step explanation:

5 0

3 years ago

Boat A is 20 km from port on a bearing of 025° and boat B is 25 km from port on a bearing of 070°. Boat B is in distress. What b

tiny-mole [99]

Answer: Boat A should travel 152.8° to reach B

Step-by-step explanation:

The diagram illustrating the scenario is shown in the attached photo. Triangle ABP is formed. A represents the position of boat A. B represents the position of boat B. P represents the position of the port.

We would determine AB by applying the law of cosines

AB² = AP² + BP² - 2AP×BPCosP

AB² = 20² + 25² - 2 × 20 × 25 × Cos45

AB² = 1025 - 707.10678 = 317.89322

AB = √317.89322 = 17.83

We would determine the bearing of B from A by finding angle A. We would apply the sine rule.

AB/SinP = AP/Sin A

17.83/Sin45 = 20/SinA

Cross multiplying, it becomes

17.83 × SinA = 20Sin45 = 14.14

SinA = 14.14/17.83 = 0.79

A = Sin^-1(0.79) = 52.2°

The total angle at A is 65 + 52.2 = 117.2°

The angle formed outside the third quadrant is 117.2 - 90 = 27.2°

Therefore, bearing B from A is

180 - 27/2 = 152.8°

3 0

3 years ago

What is the area of ABC?

Salsk061 [2.6K]

Answer:

Area of triangle ABC = 10

Option C is correct.

Step-by-step explanation:

We need to find area of triangle ABC

We are given A(2,1) B(4,7) and C(6,3)

The formula used to find $Area \ of \ triangle=\frac{1}{2}base\times height$ ABC is:

In the given triangle base= AC and height = BC

First we need to find the distance between A and C and B and C

The formula used to find distance between AC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=2, y_1=1, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-2)^2+(3-1)^2}\\Distance=\sqrt{(4)^2+(2)^2}\\Distance=\sqrt{16+4}\\Distance=\sqrt{20}$

find distance between BC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=4, y_1=7, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-4)^2+(3-7)^2}\\Distance=\sqrt{(2)^2+(4)^2}\\Distance=\sqrt{4+16}\\Distance=\sqrt{20}$

So, We have height = $\sqrt{20}$ and base = $\sqrt{20}$

Finding area of triangle

$Area \ of \ triangle=\frac{1}{2}base\times height\\Area \ of \ triangle=\frac{1}{2}\sqrt{20} \times \sqrt{20} \\Area \ of \ triangle=\frac{1}{2}\times 20\\Area \ of \ triangle=10$