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2 years ago

5

Please help !! Geometry homework. Will mark brainliest answer !

1 answer:

sergeinik [125]2 years ago

3 0

For this case we have:

By trigonometric property we have:

$Cosine(\alpha)=\frac{AdjacentLeg}{Hypotenuse}$

Where:

$Adjacent Leg = 12\\Hypotenuse = 13\\\alpha= x$

Substituting:

$Cosine(x)=\frac{12}{13}$

Clearing x we have:

$x =Cosine^{-1}(\frac{12}{13})$

Thus, the equation is: $x =Cosine^{-1}(\frac{12}{13})$

Answer:

$x =Cosine^{-1}(\frac{12}{13})$

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Some body can help me with a geometric mean maze

Mars2501 [29]

Answer:

See explanation

Step-by-step explanation:

Theorem 1: The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Theorem 2: The length of each leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the segment of the hypotenuse adjacent to that leg.

1. Start point: By the 1st theorem,

$x^2=25\cdot (49-25)=25\cdot 24=5^2\cdot 2^2\cdot 6\Rightarrow x=5\cdot 2\cdot \sqrt{6}=10\sqrt{6}.$

2. South-East point from the Start: By the 2nd theorem,

$x^2=40\cdot (40+5)=4\cdot 5\cdot 2\cdot 9\cdot 5\Rightarrow x=2\cdot 5\cdot 3\cdot \sqrt{2}=30\sqrt{2}.$

3. West point from the previous: By the 2nd theorem,

$x^2=(32-20)\cdot 32=4\cdot 3\cdot 16\cdot 2\Rightarrow x=2\cdot 4\cdot \sqrt{6}=8\sqrt{6}.$

4. West point from the previous: By the 1st theorem,

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5. West point from the previous: By the 2nd theorem,

$10^2=8\cdot (8+x)\Rightarrow 8+x=12.5,\ x=4.5.$

6. North point from the previous: By the 1st theorem,

$x^2=48\cdot 6=6\cdot 4\cdot 2\cdot 6\Rightarrow x=6\cdot 2\cdot \sqrt{2}=12\sqrt{2}.$

7. East point from the previous: By the 2nd theorem,

$x^2=22.5\cdot 30=225\cdot 3\Rightarrow x=15\sqrt{3}.$

8. North point from the previous: By the 1st theorem,

$x^2=7.5\cdot 36=270\Rightarrow x=3\sqrt{30}.$

8. West point from the previous: By the 2nd theorem,

$x^2=12.5\cdot (12.5+13.5)=12.5\cdot 26=25\cdot 13\Rightarrow x=5\sqrt{13}.$

9. North point from the previous: By the 1st theorem,

$12^2=x\cdot 30\Rightarrow x=\dfrac{144}{30}=4.8.$

101. East point from the previous: By the 1st theorem,

$6^2=1.6\cdot (x-1.6)\Rightarrow x-1.6=22.5,\ x=24.1.$

11. East point from the previous: By the 2nd theorem,

$20^2=32\cdot (32-x)\Rightarrow 32-x=12.5,\ x=19.5.$

12. South-east point from the previous: By the 2nd theorem,

$18^2=x\cdot 21.6\Rightarrow x=15.$

13. North point=The end.

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