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3 years ago

13

Lindsay draws a right triangle and adds the measures of the right angle and one acute angle. Which is a possible sum of the two

angles?

2 answers:

Anon25 [30]3 years ago

6 0

Anything between 91 degrees and 179 degrees. You can pick any between these two.

Viktor [21]3 years ago

4 0

Answer:

$90< \gamma< 180$

Step-by-step explanation:

An acute angle is an angle that measures more than 0º and less than 90º. And a right angle is an angle that measures exactly 90º. Thus:

Let:

$\alpha = Right\hspace{3}angle=90^{\circ}\\\beta= Acute\hspace{3}angle\hspace{10}0^{\circ}$

So, in this sense, the sum of the two angles can´t be equal or greater than 180º, since $\beta$ , also it has to be at least greater than 90º since $\beta>0^{\circ}$

Therefore the sum of the two angles is:

$90^{\circ}< \gamma< 180^{\circ}$

In another words, the sum is greater than 90º and less than 180º

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Cual es el 20% de 75

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In the △ABC, the height AN = 24 in, BN = 18 in, AC = 40 in. Find AB and BC. Consider all possible cases. Case 1 : N∈BC, AB = __,

aivan3 [116]

Answer:

Case 1:

$AB = 30$

$BC = 50$

Case 2:

$AB = 15.9$

$BC = 36.7$

Case 3: Not possible

Step-by-step explanation:

Given

See attachment for illustration of each case

Required

Find AB and BC

Case 1:

Using Pythagoras theorem in ANB, we have:

$AB^2 = AN^2 + BN^2$

This gives:

$AB^2 = 24^2 + 18^2$

$AB^2 = 576 + 324$

$AB^2 = 900$

Take square roots of both sides

$AB = \sqrt{900$

$AB = 30$

To calculate BC, we consider ANC, where:

$AC^2 = AN^2 + NC^2$

$40^2 = 24^2 + NC^2$

$1600 = 576 + NC^2$

Collect like terms

$NC^2 = 1600 - 576$

$NC^2 = 1024$

Take square roots

$NC = \sqrt{1024$

$NC = 32$

So:

$BC = NC + BN$

$BC = 32 + 18$

$BC = 50$

Case 2:

Using Pythagoras theorem in ANB, we have:

$AN^2 = AB^2 + BN^2$

This gives:

$24^2 = AB^2 + 18^2$

$576 = AB^2 + 324$

Collect like terms

$AB^2 = 576 - 324$

$AB^2 = 252$

Take square roots of both sides

$AB = \sqrt{252$

$AB = 15.9$

To calculate BC, we consider ABC, where:

$AC^2 = AB^2 + BC^2$

$40^2 = 252 + BC^2$

$1600 = 252 + BC^2$

Collect like terms

$BC^2 = 1600 - 252$

$BC^2 = 1348$

Take square roots

$BC = \sqrt{1348$

$BC = 36.7$

Case 3:

This is not possible because in ANC

The hypotenuse AN (24) is less than AC (40)

3 0

3 years ago

Please explain step by step for 0.67÷0.3=?

pentagon [3]

Multiply each of the numbers by a ten to a power that they both become whole numbers.

0.67 * $10^2$ = 67
0.3 * $10^2$ = 30

Divide as usual.

67 / 30 = 2 $\frac{1}{3}$

So, the answer is 2 1/3.

3 0

3 years ago

Can someone check if this is right and correct me if it’s not????

mylen [45]

Answer:

No. it is not tangent to the circle ⇒ A

Step-by-step explanation:

In any circle a tangent and a diameter are perpendicular at one end of the diameter

In the attached figure

∵ The line segment drawn from point A is passing through

the center of the circle

∴ This line segment is a diameter of the circle

∵ If AB is a tangent to the circle at point A

- That means the diameter and the tangent meet each other

at point A (one end of the diameter)

∴ AB and the diameter are perpendicular at point A

∴ m∠A = 90°

Now let us check if the lengths of the triangle formed a right angle triangle or not by using the inverse of Pythagoras Theorem

<em>In a triangle if the square of the longest side is equal to the sum of the squares of the other two sides, then the triangle is a right triangle at the vertex which opposite to the longest side</em>

∵ 17 is the longest side in the triangle

∵ The other two sides are 10.2 units and 6.8 units

∵ (17)² = 289

∵ (10.2)² + (6.8)² = 150.28

∴ (17)² ≠ (10.2)² + (6.8)²

∴ The triangle can not be a right triangle because the square

of the longest side is not equal to the sum of the squares of the

other two sides

∵ ∠A is opposite to the longest side

∴ m∠A ≠ 90°

∴ Line AB is not perpendicular to the diameter at A

∴ Line AB is not a tangent to the circle

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3 years ago