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2 years ago

Need help ASAP!!!!!

Mathematics

1 answer:

Mashutka [201]2 years ago

3 0

<span>The distance from the vertex to the centroid is twice the distance from the centroid to the side.
This means that 3x + 5 = 2 * (2x)

</span><span>3x + 5 = 4x

x = 5

Source:
http://www.1728.org/trictr.htm

</span>

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Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

3 0

2 years ago

Read 2 more answers

tamaranim1 [39]

Answer:

3000

Step-by-step explanation:

640x3 = 1920

90x12 = 1080

7 0

1 year ago

We are given arc KL is 12x+4, arc KJ is 8x-6, and angle E is 99 degrees. What is the actual degree measure of arc KJ? *

Goryan [66]

Answer:

I think its 20x - 2 because when u are given KJ or KL we need the measurement of x

4 0

2 years ago

Seven less than sixteen times a number is equal to the same number reduced by 5. What is the number?

Romashka [77]

Answer:

x = 2/15

Step-by-step explanation:

first, create a formula or expression

16x - 7 = x - 5

move x, subtract it

15x - 7 = -5

add 7

15x = 2

divide by 15

x = 2/15

sub in to check:

-4.85667 = -4.86667

7 0

2 years ago

Zaheer, a boy of height 1.5m was watching the entire programme. Initially he observed the top of the flag pole at an angle of el

7nadin3 [17]

Answer:

15.2 m

Step-by-step explanation:

You need to draw a figure. Start by drawing a horizontal segment approximately 10 cm long; that is the ground. Label the left end point A and the right endpoint B. On the right endpoint, B, go up a short 1 cm vertically. That is 1.5 m, the height of Zaheer. Label that point C. Now from that point draw a horizontal line that ends up above point A. Label that point D. Now go back to point C. Draw a segment up to the left at a 30 deg angle with CD. End the segment vertically above point D. Label that point E. That is the top of the flagpole. Draw a vertical segment down from point E through point D ending at point A. Segment AE is the flagpole. Go back to point C. Move 3 cm to the left on segment CD, and draw a point there and label it F. That is where Zaheer moved to. Now connect point F to point E. That is a 45-deg elevation to point E, the top of the flagpole.

m<EFD = 45 deg

m<EFC = 135 deg

m<FEC = 15 deg

m<ECD = 30 deg

We now use the law of sines to find EC

(sin 15)/10 = (sin 135)/EC

EC = 27.32

Because of the 30-60-90 triangle, ED = EC/2

ED = 13.66

Now we add the height of Zaheer to find AE.

13.66 + 1.5 = 15.16

Answer: 15.2 m

7 0

2 years ago