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3 years ago

15

Geometry, Triangle in a circle. Find X show work and explain please

1 answer:

nikklg [1K]3 years ago

7 0

Answer: $x=7$

Step-by-step explanation:

By the Intersecting Secants Theorem, we know that:

$(5)(x+5)=(6)(6+4)$

Having this, we can find the value of "x" by solving for "x":

Applying Distributive property:

$(5)(x+5)=(6)(24)\\5x+25=60$

Subtract 25 from both sides of the equation:

$5x+25-25=60-25\\5x=35$

And finally dviding both sides of the equation by 5, we get:

$\frac{5x}{5}=\frac{35}{5}\\x=7$

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3 years ago

Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3> (1 point)

sesenic [268]

Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

<u>Step-by-step explanation:</u>

We have , two vectors u = <-5, -4>, v = <-4, -3> or , $u = -5i-4j , v=-4i-3j$

We need to find angle between these two vectors . Let's find out:

We know that dot product of two vectors is defined as :

$u.v =|u|(|v|)cosx$ , where x is angle between u & v !

⇒ $u.v =|u|(|v|)cosx$

⇒ $cosx =\frac{u.v}{|u|(|v|)}$

Now , $u.v = (-5i-4j)(-4i-3j)$

⇒ $u.v = (-5i-4j)(-4i)-(-5i-4j)(3j)$

⇒ $u.v = 20+12$ { $i(j) = j(i) =0$ }

⇒ $u.v = 32$

Now , Modulus of any vector $r = xi+yj$ is $|r| = \sqrt{x^{2}+y^{2}}$ So ,

$|u| = \sqrt{(-5)^{2}+(-4)^{2}} = \sqrt{25+16} = \sqrt{41} \\\\|v| = \sqrt{(-4)^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5$

Putting all these values in equation $cosx =\frac{u.v}{|u|(|v|)}$ we get:

⇒ $cosx =\frac{32}{5(\sqrt{41})}$

⇒ $cos^{-1}(cosx) =cos^{-1}(\frac{32}{5(6.4)})$

⇒ $x =cos^{-1}(1)$ { $cos0 = 1$ }

⇒ $x =0$ °

Therefore , Angle between $u = -5i-4j , v=-4i-3j$ is $x =0$ ° .

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