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Dvinal [7]
2 years ago
7

Given: △ABC, m∠B=90° AB=12, BC=16, BK ⊥ AC . Find: AC and BK. Please explain how to find out BK​

Mathematics
2 answers:
Kryger [21]2 years ago
7 0

Answer:

Given: △ABC, m∠B=90° AB=12, BC=16, BK ⊥ AC . Find: AC and BK.

Given: △ABC, m∠B=90°

Find: AC and BK.

Short leg 90 degrees Long leg Hypotenuse

AB=12 90 BC=16 AC= ?

AK = ? 90 BK = ? AB=12

AC = SQRT (AB*AB + BC*BC) = 20 [right triangle; Pythagorean Theorem]

Similar triangles:[Note: In diagram, share two angles. Therefore share three angles]

BK / 16 = AB / AC

BK / 16 = 12 / 20

BK = (3/5)16

BK = 48/5

another answer let see this

AB^2+BC^2=AC^2

12^2+16^2=AC^2

144+256=AC^2

400=AC^2

20=AC

# be careful#

goldenfox [79]2 years ago
6 0

Answer: BK = 9.6

Step-by-step explanation: The equation of the line AC is y = (-3/4)x + 12. The line perpendicular to AC and passing through point B is y = (4/3)x. The point of intersection between BK amd AC is (144/25, 192/25). Those are the side lengths. Use Pythagorean theorem to find the hypotenuse.

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Y = 3/5x + 1, 5y = 3x - 2, 10x - 6y = -4<br> is it perpendicular, parallel, neither
Ber [7]

Answer:

y=\frac{3}{5} x+1 and 5y=3x-2 are parallel.

10x-6y=-4 is neither parallel nor perpendicular.

Step-by-step explanation:

First, you have to simplify each equation in terms of y.

y=\frac{3}{5} x+1\\5y=3x-2\\10x-6y=-4

Your first equation is already in terms of x, so simplify your second equation.

5y=3x-2\\y=\frac{3}{5} x-\frac{2}{5}

Now you can simplify your third equation.

10x-6y=-4\\-6y=-10x-4\\y=\frac{5}{3} x+\frac{2}{3}

These are your three equations in terms of y:

y=\frac{3}{5} x+1\\\\y=\frac{3}{5} x-\frac{2}{5} \\\\y=\frac{5}{3} x+\frac{2}{3}

Now, all you have to know is how to tell using your slope if a line is parallel or perpendicular to another.

Two parallel lines will have the exact same slope.

Two perpendicular lines will have slopes which are opposite reciprocals. For example, a line with a slope of 2 is perpendicular to a line with a slope of -\frac{1}{2}, as they have opposite signs and are reciprocal (2/1 versus 1/2) to each other.

Your first two equations have the same slope and are therefore parallel.

Your third equation is a reciprocal, but it is not opposite, and is therefore not parallel nor perpendicular.

5 0
3 years ago
PLEASE HELP ME IM STRUGGLING!!!
Kitty [74]

Answer:

The required answer is c=7\sqrt{3}

Therefore the number in green box should be 7.

Step-by-step explanation:

Given:

AB = 7√2

AD = a , BD = b , DC = c , AC = d

∠B = 45°, ∠C = 30°

To Find:

c = ?

Solution:

In Right Angle Triangle ABD Sine identity we have

\sin B = \dfrac{\textrm{side opposite to angle B}}{Hypotenuse}\\

Substituting the values we get

\sin 45 = \dfrac{AD}{AB}= \dfrac{a}{7\sqrt{2}}

\dfrac{1}{\sqrt{2}}= \dfrac{a}{7\sqrt{2}}\\\\\therefore a=7

Now in Triangle ADC Tangent identity we have

\tan C = \dfrac{\textrm{side opposite to angle C}}{\textrm{side adjacent to angle C}}

Substituting the values we get

\tan 30 = \dfrac{AD}{DC}= \dfrac{a}{c}\\\\\dfrac{1}{\sqrt{3}}=\dfrac{7}{c}\\\\\therefore c=7\sqrt{3}

The required answer is c=7\sqrt{3}

8 0
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Step-by-step explanation:

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