Question

100 POINTS + BRAINLIST

Answer 1

Problem 1

Since point P is the tangent point, this means angle OPT is a right angle

angle OPT = 90 degrees

Let's use the Pythagorean theorem to find the missing side 'a'

a^2 + b^2 = c^2

a^2 + 7^2 = 12^2

a^2 + 49 = 144

a^2 = 144 - 49

a^2 = 95

a = sqrt(95)

a = 9.7467943

a = 9.7

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Let's use the sine and arcsine rule to find angle y

sin(angle) = opposite/hypotenuse

sin(y) = 7/12

y = arcsin( 7/12 )

y = 35.6853347

y = 35.7

The value of y is approximate. Make sure your calculator is in degree mode. Arcsine is the same as inverse sine or $\sin^{-1}$

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Problem 2

Focus on triangle OHP. This may or may not be a right triangle. The goal is to test if it is or not.

We use the converse of the Pythagorean theorem to check.

Recall that the converse of the Pythagorean theorem says: If a^2+b^2 = c^2 is a true equation, then the triangle is a right triangle. The value of c is always the longest side. The order of a and b doesn't matter.

In this case we have

• a = 7
• b = 13
• c = 16

Which leads to

a^2 + b^2 = c^2

7^2 + 13^2 = 16^2

49 + 169 = 256

218 = 256

We get a false equation at the end, since both sides aren't the same number, which means the original equation is false.

We don't get a^2 + b^2 = c^2 to be true, therefore this triangle is not a right triangle.

Consequently, this means angle OPH cannot possible be 90 degrees (if it was then we'd have a right triangle). Therefore, point P is not a tangent point.

You follow the same basic idea for triangle OQH and show that point Q is not a tangent point either. Or you could use a symmetry argument to note that triangle OPH is a mirror reflection of triangle OQH over the line segment OH. This implies that whatever properties triangle OPH has, then triangle OQH has them as well for the corresponding pieces.

Answer 2

Answer to Question 1:

y˚ = 36˚

a = 9.7

Step-by-step explanation:

How to find the value of a

When there is a line that’s tangent of a circle, it’s ALWAYS going to be perpendicular to the line of the centre of the circle. So you can see that line PT is perpendicular to the line that comes out from the centre of the circle (the line that’s labeled as PO, or labeled as a length of 7). This means that it forms a right angle (two lines that are perpendicular to each other).

Like the other problem I solved for you, you can use the Pythagorean Theorem ($a^{2}+b^{2}=c^{2}$) to find what a is, because you can see that it forms a triangle in the diagram. But here is one main thing to know when using the Pythagorean Theorem: The triangle HAS TO BE a right triangle, meaning that it has to have a right angle. But we already proved that it forms a right angle because line PT is perpendicular/tangent to line PO (labeled as a length of 7). So, we can for sure use the Pythagorean Theorem. The steps are shown below:

Steps (the variables a, b, and c are in italics so it’s not confusing to look at):

-You can rearrange the formula from the Pythagorean Theorem, which is $a^{2}+b^{2}=c^{2}$ to only get a because we want to know what a is.

Rearranged formula: $a=\sqrt{c^{2} -b^{2} }$  or  $a=-\sqrt{c^{2} -b^{2} }$  It has to ways/options for a to equal because when finding the square root of a number, it can be either positive or negative. But in this problem, it’s going to be positive because we’re finding the length of something.

-The variable c in the formula means the hypotenuse of the triangle. In the diagram, you can see that 12 is the hypotenuse, which means that b is 7.

-Since we know what c and b are, you can plug them into the new formula:

$a=\sqrt{12^{2}-7^{2} }$

-Simplify it:

$a=\sqrt{144-49}$

-Subtract inside the radical:

$a=\sqrt{95}$  Then simplify it to get: $a= 9.74679434...$  And it wants us to round to the nearest tenth. So you finally get $a=9.7$.

You can check your answer by putting a back into the original Pythagorean Theorem formula ($a^{2} +b^{2}= c^{2}$):

Steps

-Plug a, b, and c into the formula (use $\sqrt{95}$ for a because using a rounded decimal won’t give you an exact answer):

$(\sqrt{95^{2}})+(7^{2})=(12^{2})$

-Simplify to get:

$95+49=144$

-Add:

$144=144$

This is very true. So our answer is correct.

ANSWER: $a=9.7$

How to solve for y˚

Since all three angles in a triangle add up to 180˚ and you already know two of them, you should know what the third one is. The two angles that are given are 90˚ (because it’s a right triangle) and 54˚.

You can be written as:

90˚ + 54˚ + y = 180˚

Solve for y:

144˚ + y = 180˚

Subtract 134˚ from both sides to get y by itself:

y = 36˚

ANSWER: y˚ = 36˚

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Answer to Question 2:

Reminder for you: If you don’t remember from the very beginning on the top of the the page, I said that inorder to know if a line is tangent, it’s ALWAYS going to be perpendicular to the line of the centre of the circle.

So we need to test if line PH is perpendicular to the line that comes out from the centre of the circle (the line that’s labeled as PO, or labeled as a length of 7 cm).

We can do this by using the Pythagorean Theorem ($a^{2}+b^{2}=c^{2}$) because if you still don’t remember what I said at the very beginning, I said that there is one main thing to know when using the Pythagorean Theorem: The triangle HAS TO BE a right triangle, meaning that it has to have a right angle.

So we can use the Pythagorean Theorem to find if line PH and line PO (labeled as 7 cm) are perpendicular/form a right angle.

Here are your mathematical calculations of the answer:

-Plug the values from the numbers that are from the triangle:

$13^{2} +7^{2} =16^{2}$  (16 is the hypotenuse, c, because it’s the longest side of the triangle)

-Simplify:

$169+49=256$

-Add:

$218=256$

This is false because 218 does not equal 256. Therefore, the triangle does not have a right angle, which means that line PH is not perpendicular/tangent to line PO.

ANSWER: The wire is not a tangent to the circle at P and at Q because when using the Pythagorean Theorem to prove if line PH is perpendicular to line PO, the result is false.

Sorry for the very, very, long explanation. But hopefully you understand and that this helps with your question! :)