A rectangle has dimensions 10 cm by 6 cm. determine the measures of the angles at the point of where the diagonals intersect

Mathematics

1 answer:

Sophie [7]3 years ago

3 0

See the attached figure to better understand the problem

we know that

in the triangle ABC
tan(alfa/2)=AC/BC----> 5/3
so
(alfa/2)=arctan (5/3)---------> 59.0362°
alfa=2*59.0362--------> alfa=118.07°
alfa+beta=180°----------> because are supplementary angles
beta=180-118.07-------> beta=61.93°

the answer is
the angles are
alfa=118.07°
beta=61.93°

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The length of a triangel is three times its with.if the perimiter is at its most 112 centemeters,witch is the greates possible v

iris [78.8K]

I am assuming you do not mean The length of a triangle and you mean The length of a rectangle because the answer you given are the formula for the perimeter of a rectangle. With that said, the answer is D<span>.2w+2*(3w)112

The perimeter of a rectangle can be found by the following equation:

2L + 2W = P
OR
P = 2L + 2W
The equation </span>P = 2L + 2W is read perimeter = 2 times Length + 2 times width

The Question:
<span>The length of a rectangle is three times its width.if the perimeter is at its most 112 centimeters ,which is the greatest possible value of the the width.
</span>
The length of a rectangle
This means the perimeter

is
The word is means to use an equal sign

three times its width
This means 3 times w or 3w
The w = width

perimeter is at its most 112
This means <=
<= means less than or equal to

Ok so lets look at our equation

2L + 2W = P
2w + 2(3w) <= 112

5 0

3 years ago

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$