Question

Solve using law of sines or law of cosines!

Answer 1

Answer:

Part 5) The length of the ski lift is $1.15\ miles$

Part 6) The height of the tree is 18.12 m

Step-by-step explanation:

Part 5)

Let

A -----> Beginning of the ski lift

B -----> Top of the mountain

C -----> Base of mountain

we have

$b=0.75\ miles$

$A=20\°$

$C=180\°-50\°=130\°$ ----> by supplementary angles

Find the measure of angle B

Remember that the sum of the interior angles must be equal to 180 degrees

$B=180\°-A-C$

substitute

$B=180\°-20\°-130\°=30\°$

Applying the law of sines

$\frac{b}{sin(B)}=\frac{c}{sin(C)}$

substitute

$\frac{0.75}{sin(30\°)}=\frac{c}{sin(130\°)}$

$c=\frac{0.75}{sin(30\°)}(sin(130\°))$

$c=1.15\ miles$

Par 6)

see the attached figure with letters to better understand the problem

Applying the law of sines in the right triangle BDC

In the right  triangle BDC 20 degrees is the complement of 70 degrees

$\frac{BC}{sin(70\°)}=\frac{x}{sin(20\°)}$

$BC=(sin(70\°))\frac{x}{sin(20\°)}$ -----> equation A

Applying the law of sines in the right triangle ABC

In the right  triangle ABC 50 degrees is the complement of 40 degrees

$\frac{BC}{sin(40\°)}=\frac{x+15}{sin(50\°)}$

$BC=(sin(40\°))\frac{x+15}{sin(50\°)}$ -----> equation B

Equate equation A and equation B and solve for x

$(sin(70\°))\frac{x}{sin(20\°)}=(sin(40\°))\frac{x+15}{sin(50\°)}\\\\2.7475x=0.8391(x+15)\\\\2.7475x=0.8391x+12.5865\\\\2.7475x-0.8391x=12.5865\\\\x=6.60\ m$

Find the value of BC

$BC=(sin(70\°))\frac{6.6}{sin(20\°)}$

$BC=18.12\ m$

therefore

The height of the tree is 18.12 m