Question

Finding Hypotenuse Lengths. Find the length of the hypotenuse.

Answer 1

Answer:

11) $\frac{3\sqrt{5}}{2}$

12) $4\sqrt{5}$

13) $22$

14) $18$

15) $2$

16) $24\sqrt{3}$

Step-by-step explanation:

For these problems I used the pythagorean theorem: $a^{2}+ b^{2}= c^{2}$ and SOHCAHTOA

Sin = $\frac{opposite}{hypotenuse}$

Cos = $\frac{adjacent}{hypotenuse}$

Tan = $\frac{opposite}{hypotenuse}$

11)

First find the length of the bottom side by using Cos

$cos(60)=\frac{x}{3}$

$3(cos(60))=x$

$1.5=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$3^{2}+ 1.5^{2}= c^{2}$

$9+2.25=c^{2}$

$\sqrt{11.25}=\sqrt{c^{2}}$

$\frac{3\sqrt{5}}{2}$

12)

Find the length of the bottom side using Cos

$cos(60)=\frac{x}{8}$

$8(cos(60))=x$

$4=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$8^{2}+ 4^{2}= c^{2}$

$64+16=c^{2}$

$\sqrt{80} =\sqrt{c^{2} }$

$4\sqrt{5}$

13)

Find the length of the other side by using Tan

$tan(30)=\frac{x}{11\sqrt{3} }$

$11\sqrt{3}* (tan(30)=x$

$11=x$

Then plug it into the formula for the pythagorean theorem to find the hypotenuse

$(11\sqrt{3}) ^{2}+ 11^{2}= c^{2}$

$363+121=c^{2}$

$\sqrt{484} =\sqrt{c^{2}}$

$22$

14)

(This is probably an easier way to do these problems)

Find the hypotenuse by using Cos ($\frac{adjacent}{hypotenuse}$)

$cos(60)=\frac{9}{x}$

$cos(60)x=9$

$x=\frac{9}{cos(60)}$

$x=18$

15)

Find the hypotenuse using Sin ($\frac{opposite}{hypotenuse}$)

$sin(30)=\frac{1}{x}$

$sin(30)x=1\\x=\frac{1}{sin(x)}$

$x=2$

16)

Find the hypotenuse using Cos ($\frac{adjacent}{hypotenuse}$)

$cos(60)=\frac{12\sqrt{3} }{x}$

$cos(60)x=12\sqrt{3}$

$x=\frac{12\sqrt{3} }{cos(60)}$

$x=24\sqrt{3}$