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3 years ago

If ABCD is a rectangle, calculate x as a function of α

Mathematics

2 answers:

shusha [124]3 years ago

7 0

Answer:

x = 90 - 2α

Step-by-step explanation:

Solution:-

- Consider the right angled triangle " ABD ". The sum of angles of an triangle is always "180°".

< BAD > + < ADB > + < ABD > = 180°

< ABD > = 180 - 90° - α

< ABD > = 90° - α

- Then we look at the figure for the triangle "ABE". Where " E " is the midpoint and intersection point of two diagonals " AC and BD ".

- We name the foot of the perpendicular bisector as " F ": " BF " would be the perpendicular bisector. The angle < BAE > is equal to < ABD >.

< ABD > = < BAE > = 90° - α ... ( Isosceles triangle " BEA " )

Where, sides ( BE = AE ).

- Use the law of sum of angles in a triangle and consider the triangle " BFA " as follows:

< ABF> + < BFA > + < BAF > = 180°

< ABF > = 180 - (90° - α) - 90°

< ABF > = α

Where, < BAF > = < BAE >

- The angle < ABD > = < ABE > is comprised of two angles namely, < ABF > and < FBE > = x.

< ABD > = < ABE > = < ABF > + x

90° - α = α + x

x = 90 - 2α ... Answer

Yanka [14]3 years ago

4 0

Answer:

Step-by-step explanation:

The length of this triangle is 10 squares

and the width is 4 squares

The diagonals divide the rectangle into four triangles

These traingles are isoceles

Each two triangles facing each others are identical

<B = 90 degree

B = alpha + Beta

Let Beta be the angle next alpha

The segment that is crossing Beta is its bisector since it perpendicular to the diagonals wich means that:

Beta = 2x

Then B = alpha + 2x

90 = alpha +2x

90-alpha = 2x

x = (90-alpha)/2

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3 years ago

Please help and show work y = 7x - 3 - 4x y = -3 + 3x

eimsori [14]

Answer:

y = 3 x − 3

Step-by-step explanation:

Subtract 4 x from 7 x

= y = 3 x − 3

3 0

3 years ago

What’s the ratios of each

bagirrra123 [75]

Given:

The triangle ABC is a right triangle.

The length of AC = 25, the length of AB = 7 and the length of BC = 24

We need to determine the ratios of sin C, cos C and tan C.

Ratio of sin C:

Using the trigonometric ratio, the ratio of sin C is given by

$sin \ C=\frac{opp}{hyp}$

where $opp=AB$ and $hyp=AC$

Thus, we get;

$sin \ C=\frac{AB}{AC}$

Substituting the values, we get;

$sin \ C=\frac{7}{25}$

Thus, the ratio of sin C is $\frac{7}{25}$

Ratio of cos C:

The ratio of cos C can be determined using the trigonometric ratio.

Thus, we have;

$cos C=\frac{adj}{hyp}$

where $adj=BC$ and $hyp=AC$

$cos \ C=\frac{BC}{AC}$

Substituting the values, we get;

$cos \ C=\frac{24}{25}$

Thus, the ratio of cos C is $\frac{24}{25}$

Ratio of tan C:

The ratio of tan C can be determined using the trigonometric ratio.

Thus, we have;

$tan \ C=\frac{opp}{adj}$

where $opp=AB$ and $adj=BC$

Thus, we have;

$tan \ C=\frac{AB}{BC}$

Substituting the values, we get;

$tan \ C=\frac{7}{25}$

Thus, the ratio of tan C is $\frac{7}{25}$

3 0

3 years ago

How many milliliters of a 15% saline solution and how many milliliters of a 40% saline solution must be mixed to produce 10 mill

Lesechka [4]

X - volume of 15% saline solution. Theres is 15% x = 0,15 x saline solution
y - volume of 40% saline solution. There is 40% y = 0,4y saline solution

You want to produce 10 mililiers so x+y= 10

And this saline has to be 30%, so you've got 30 % * 10 = 0,3 * 10 = 3 ml saline. So:
0,15x + 0,4y = 3
And you've got system of equations:
$\begin{cases} x+y=10 \\ 0,15x+0,4y=3\end{cases} \\ \begin{cases} x=10-y \\ 0,15x+0,4y=3\end{cases} \\ \hbox{Substitute value from first equation to second:} \\ 0,15(10-y)+0,4y=3 \\ 1,5-0,15y+0,4y=3 \\ 0,25y=1,5 \qquad /:0,25 \\ y=\frac{1,5}{0,25}=\frac{150}{25}=6 \\ \hbox{Then:} \\ x=10-y=10-6=4$

So answer:
Volume of 15% saline solution : 4 mililiters
Volume of 40% saline solution: 6 mililiters.

4 0

3 years ago

What is the solution set for x in log4(x+5)+log4x=log4 14

marishachu [46]

$log_ {4} (x+5) + log_4(x) = log_4(14)$

For logs with same base :

$log_b(f(x)) + log_b(g(x)) = log_b(f(x)*g(x))$

$Log_4(x+5) + log_4(x) =log_4((x+5)x)$

$log_4((x+5) = log_4(14)$

When the logs have the same base :

$log_b(f(x)) = log_b(g(x))$

$F(x) = g(x)$

For $log_4((x+5)x) = log_4(14)$

Solve $(x+5)x = 14$

expand

$(x+5)x = x^2 + 5x$

$x^2+5x+14$

$(x-2)(x+7) = 0$

$x = 2 , x = - 7$

Therefore, the final solution : $x = 2$

hope this helps!

3 0

3 years ago