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Triss [41]
3 years ago
13

Triangle DEF has vertices D(0,0), E(7,0), and F(3,7).

Mathematics
1 answer:
Leni [432]3 years ago
4 0

Answer:

(3, 1.7)

Step-by-step explanation:

The point at which the vertices of a triangle meet is known as the orthocenter of the triangle. The orthocenter passes through the vertex of the triangle and is perpendicular to the opposite sides.

Two lines are perpendicular if the product of their slopes is -1.

The slope of the line joining D(0,0), F(3,7) is:

m_1=\frac{7-0}{3-0}=\frac{7}{3}

The slope of the line perpendicular to the line joining D and F is -3/7. The orthocenter is perpendicular to the line joining D and F and passes through vertex E(7, 0). The equation is hence:

y-y_1=m(x-x_1)\\\\y-0=-\frac{3}{7} (x-7)\\\\y=-\frac{3}{7}x+3 \ .\ .\ .\ (1)

The slope of the line joining E(7,0), and F(3,7). is:

m_1=\frac{7-0}{3-7}=-\frac{7}{4}

The slope of the line perpendicular to the line joining E and F is 4/7. The orthocenter is perpendicular to the line joining E and F and passes through vertex D(0, 0). The equation is hence:

y-y_1=m(x-x_1)\\\\y-0=\frac{4}{7} (x-0)\\\\y=\frac{4}{7} x\ .\ .\ .\ (2)

The point of intersection of equation 1 and equation 2 is the orthocenter. Solving equation 1 and 2 simultaneously gives:

x = 3, y = 1.7

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4 years ago
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Answer:

A=28.1\ cm^2

Step-by-step explanation:

The area of a circular sector with a central angle θ and radius r is given by:

\displaystyle A=\frac{1}{2}r^2 \theta

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We need to find both the radius and the angle in radians.

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Radius r=Diameter/2=4.8 cm

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Sin(A+B) sin(A-B) /sin^A Cos^B=1-cot^A Tan^B​
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In order to prove

\dfrac{\sin(x+y)\sin(x-y)}{\sin^2(x)\cos^2(x)}=1-\cot^2(x)\tan^2(y)

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Let's start with the left hand side: we can use the formula for sum and subtraction of the sine to write

\sin(x+y)=\cos(y)\sin(x)+\cos(x)\sin(y)

and

\sin(x-y)=\cos(y)\sin(x)-\cos(x)\sin(y)

So, their multiplication is

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So, the left hand side simplifies to

\dfrac{\cos^2(y)\sin^2(x)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now, on with the right hand side. We have

1-\cot^2(x)\tan^2(y)=1-\dfrac{\cos^2(x)}{\sin^2(x)}\cdot\dfrac{\sin^2(y)}{\cos^2(y)} = 1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

Now simply make this expression one fraction:

1-\dfrac{\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}=\dfrac{\sin^2(x)\cos^2(y)-\cos^2(x)\sin^2(y)}{\sin^2(x)\cos^2(y)}

And as you can see, the two sides are equal.

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