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2 years ago

Given: △ABC, AB=BC=10 cm, AC=12 cm, AK ⊥ BC CM ⊥ AB , BN ⊥ AC . Find: AK, BN, CM.

1 answer:

7 0

Since AB=Bc, triangle ABC is isosceles with base AC. Then BN is height and median drawn to the base. If BN is median and AC=12 cm, then AN=NC=6 cm.

1. By the Pythagorean theorem for triangle ABN,

$AB^2=AN^2+BN^2,\\ \\10^2=6^2+BN^2,\\ \\BN^2=100-36=64,\\ \\BN=8\ cm.$

2. AB and BC are legs of isosceles triangle ABC. The heights drawn to the lags of isosceles triangle are congruent, so AK=CM. Consider two right triangles AKC and AKB. By the Pythagorean theorem,

$AC^2=AK^2+KC^2,\\ \\AB^2=BK^2+AK^2=(BC-KC)^2+AK^2.$

Subtract from the second equation the first one:

$AB^2-AC^2=(BK-KC)^2-KC^2,\\ \\10^2-12^2=(10-KC)^2-KC^2,\\ \\100-144=100-20KC+KC^2-KC^2,\\ \\-144=-20KC,\\ \\KC=7,2\ cm.$

Then

$AK^2=AC^2-KC^2,\\ \\AK^2=12^2-7.2^2=144-51,84=92.16,\\ \\AK=9.6\ cm.$

Answer: AK=CM=9.6 cm, BN=8 cm

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Answer:

2(b+1+b)= 18

Step-by-step explanation:

Perimeter of a rectangle= 2(l+b)

18 feet= 2(l+b)

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Because the length of the sandbox is 1 foot longer, it will be:

b+1 = l

Substitute l with b+1

2(b+1+b)= 18

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2 years ago

How is this solved using trig identities (sum/difference)?

GenaCL600 [577]

FIRST PART
We need to find sin α, cos α, and cos β, tan β
α and β is located on third quadrant, sin α, cos α, and sin β, cos β are negative

Determine ratio of ∠α
Use the help of right triangle figure to find the ratio
tan α = 5/12
side in front of the angle/ side adjacent to the angle = 5/12
Draw the figure, see image attached

Using pythagorean theorem, we find the length of the hypotenuse is 13
sin α = side in front of the angle / hypotenuse
sin α = -12/13

cos α = side adjacent to the angle / hypotenuse
cos α = -5/13

Determine ratio of ∠β
sin β = -1/2
sin β = sin 210° (third quadrant)
β = 210°

$cos \beta = -\frac{1}{2} \sqrt{3}$

$tan \beta= \frac{1}{3} \sqrt{3}$

SECOND PART
Solve the questions
Find sin (α + β)
sin (α + β) = sin α cos β + cos α sin β
$sin( \alpha + \beta )=(- \frac{12}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{5}{13} )( -\frac{1}{2} )$
$sin( \alpha + \beta )=(\frac{12}{26}\sqrt{3})+( \frac{5}{26} )$
$sin( \alpha + \beta )=(\frac{5+12\sqrt{3}}{26})$

Find cos (α - β)
cos (α - β) = cos α cos β + sin α sin β
$cos( \alpha + \beta )=(- \frac{5}{13} )( -\frac{1}{2} \sqrt{3})+( -\frac{12}{13} )( -\frac{1}{2} )$
$cos( \alpha + \beta )=(\frac{5}{26} \sqrt{3})+( \frac{12}{26} )$
$cos( \alpha + \beta )=(\frac{5\sqrt{3}+12}{26} )$

Find tan (α - β)
$tan( \alpha - \beta )= \frac{ tan \alpha-tan \beta }{1+tan \alpha tan \beta }$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5}{12}) ( \frac{1}{2} \sqrt{3})}$

Simplify the denominator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{1+(\frac{5\sqrt{3}}{24})}$
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{1}{2} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the numerator
$tan( \alpha - \beta )= \frac{ \frac{5}{12} - \frac{6}{12} \sqrt{3} }{ \frac{24+5\sqrt{3}}{24} }$
$tan( \alpha - \beta )= \frac{ \frac{5-6\sqrt{3}}{12} }{ \frac{24+5\sqrt{3}}{24} }$

Simplify the fraction
$tan( \alpha - \beta )= (\frac{5-6\sqrt{3}}{12} })({ \frac{24}{24+5\sqrt{3}})$
$tan( \alpha - \beta )= \frac{10-12\sqrt{3} }{ 24+5\sqrt{3}}$

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2 years ago

A company that teaches self-improvement seminars is holding one of its seminars in Lancaster. The company pays a flat fee of $98

alukav5142 [94]

Answer:

Step-by-step explanation:

Let x represent the number of attendees that it will take the company to break even.

The company pays a flat fee of $98 to rent a facility in which to hold each session. Additionally, for every attendee who registers, the company must spend $16 to purchase books and supplies. This means that the total cost that the company would pay for x attendees is

16x + 98

Each attendee will pay $65 for the seminar. This means that the total revenue that the company would generate from x attendees is 65x.

At the break even point,

total cost = total revenue

Therefore,

16x + 90 = 65x

65x - 16x = 98

49x = 98

x = 98/49

x = 2

It will take 2 attendees and the total expenses and revenues is

2 × 65 = $130

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2 years ago

A 20 foot ladder leaning against a wall is used to reach a window that is 17 feet above the ground. How far from the wall is the

Liono4ka [1.6K]

Answer:

The wall is 10.5 foot far from the bottom of the ladder.

Step-by-step explanation:

Given:

Height of the ladder leaning against wall (Hypotenuse) = 20 foot.

Height from where the window is above from the ground (Leg1) = 17 feet.

To find the distance of the bottom of the ladder far from the wall (Leg2) = ?

Now, by using the pythagorean theorem:

$Hypotenuse^{2} = (Leg1)^{2} + (Leg2)^{2}$

$20^{2} =17^{2}+(Leg2) ^{2}$

$400=289+(Leg2)^{2}$

$400-289=(Leg2)^{2}$

$111=(Leg2)^{2}$

by squaring both sides

$Leg2=10.535$

10.535 foot rounded nearest tenth is 10.5 foot.

Therefore, the wall is 10.5 foot far from the bottom of the ladder.

6 0

2 years ago

Explain the special pattern for sketching a parabola

Lady bird [3.3K]

Answer:

the parabola can be written as:

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first step.

find the vertex at:

x = -b/2a

the vertex will be the point (-b/2a, f(-b/2a))

now, if a is positive, then the arms of the parabola go up, if a is negative, the arms of the parabola go down.

The next step is to see if we have real roots by using the Bhaskara's equation:

$x = \frac{-b +-\sqrt{b^2 -4ac} }{2a}$

Now, draw the vertex, after that draw the values of the roots in the x-axis, and now conect the points with the general draw of the parabola.

If you do not have any real roots, you can feed into the parabola some different values of x around the vertex

for example at:

x = (-b/2a) + 1 and x = (-b/2a) - 1

those two values should give the same value of y, and now you can connect the vertex with those two points.

If you want a more exact drawing, you can add more points (like x = (-b/2a) + 3 and x = (-b/2a) - 3) and connect them, as more points you add, the best sketch you will have.

8 0

2 years ago