Use cosine rule,

cos(A)=(b^2+c^2-a^2)/(2bc)
=(10^2+12^2-6^2)/(2*10*12)
=13/15
A=29.926 degrees.................................(A)

cos(B)=(c^2+a^2-b^2)/(2ca)
=(12^2+6^2-10^2)/(2*12*6)
=5/9
B=56.251 degrees.................................(B)

cos(C)=(a^2+b^2-c^2)/(2ab)
=(6^2+10^2-12^2)/(2*6*10)
=-1/15
C=93.823 degrees.................................(C)

Check:29.926+56.251+93.823=180.0 degrees....ok

5 0

3 years ago

A ball was dropped from a height of 60m. Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped. Ho

asambeis [7]

Answer:

The ball traveled 116.25 m when it hit the ground for the fifth term

Step-by-step explanation:

This is a geometric progression exercise and what we are asked to look for is the sum of a GP.

The ball was dropped from a height of 60 m. This means that the initial height of the ball is 60 m.

First value, a = 60

Each time it hit the ground, it bounced up 1/2 (half) of the height that it dropped.

This is the common ratio, r = 1/2 = 0.5

The number of terms it hits the ground is the number of terms in the GP.

number of terms, n = 5

The distance traveled by the ball when it hit the ground for the fifth term will be modeled by the equation:

$S_n = \frac{a(1 - r^n) }{1 - r} \\S_5 = \frac{60(1 - 0.5^5) }{1 - 0.5}\\S_5 = \frac{58.125}{0.5} \\S_5 = 116.25 m$

4 0

3 years ago