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2 years ago

What is the measure of LBAC?

Mathematics

2 answers:

ikadub [295]2 years ago

7 0

Answer:

correct answer is 30 degrees

JulijaS [17]2 years ago

3 0

Answer:

Step-by-step explanation:

x + 10 = 2x - 10

x + 10 + (10 - x) = 2x - 10 + (10 - x)

20 = x

∠BAC = (20 + 10) = 30°

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Help please.........

Tanya [424]

Answer:

y = 4.5 x

Step-by-step explanation:

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3 years ago

Read 2 more answers

In triangle ΔABC, ∠C is a right angle and CD is the height to

Zina [86]

Answer:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

Step-by-step explanation:

The triangles are drawn below.

CD is perpendicular to AB as CD is height to AB.

Therefore, angles $m\angle CDB=m\angle CDA=90$ °

So, triangles ΔCBD and ΔCAD are right angled triangles.

Now, from the right angled triangle ΔABC,

$m\angle A+m\angle B =90\\\alpha+m\angle B=90\\m\angle B=90-\alpha$

From ΔCBD,

$m\angle CBD$ is same as $m\angle B$ .

So, $m\angle CBD=90-\alpha$

$m\angle BCD+m\angle BDC =90\\m\angle BCD+90-\alpha=90\\m\angle BCD=\alpha$

Now, from ΔCAD,

$m\angle CAD$ is same as $m\angle A$

So, $m\angle CAD=\alpha$

$m\angle CAD+m\angle ACD =90\\\alpha+m\angle ACD=90\\m\angle ACD=90-\alpha$

Hence, the unknown angles of both the triangles are:

$m\angle CDB=90\\m\angle CBD=90-\alpha\\m\angle BCD=\alpha\\\\m\angle CDA=90\\m\angle CAD=\alpha\\m\angle ACD=90-\alpha$

5 0

3 years ago

A cement walkway of uniform width has been built around an in-ground rectangular pool. the area of the walkway is 1344 square fe

Elza [17]

Answer:

6 feet.

Step-by-step explanation:

Dimensions of the Pool =80 feet long by 20 feet wide

Area of the walkway =1344 square feet.

If the width of the walkway=w

Length of the Larger Rectangle =80+2w
Width of the Larger Rectangle =20+2w

Area of the Walkway =Area of the Larger Rectangle - Area of the Pool

$1344=(80+2w)(20+2w)-(80*20)\\1344=80*20+160w+40w+4w^2-80*20\\4w^2+200w=1344\\4w^2+200w-1344=0\\We factorize\\4(w^2+50w-336)=0\\4(w^2-6w+56w-336)=0\\w(w-6)+56(w-6)=0\\(w-6)(w+56)=0\\w-6=0$ or $ w+56=0\\w=6$ ft or $ w=-56$ \\Therefore, the width of the walkway , w=6 feet.$