we can have a triangle and we know that there are two angles.
30º and a 45º + 105 =180º angle. applying the sides.
therefore if we have the whole triangle completed we have a constant triangle.
so it tell us we can you’se SOH CAH TOA
we can varrify what kind of triangle we are using.
for example if we were to use Sin I would do
Sin 30/ 2 = sin of 105º/ A = sin=45º/ B
if you would want to solve for sin 30/2 and sin 105, then we solve.
or Sin 30/2 and Sin 105 as well.
with your calculator use sin= 30=.5 and use sin 105º a divide by 1.5 on both sides 1.5= A /sin 105º * 1.5 answer would be 3.86. a is = to 3.86 (Not the answer)
recall d = rt, distance = rate * time.
let's say airplane A is going at a rate of "r", therefore airplane B is moving faster, at a rate of "r + 80".
now, after 3 hours, both planes have been travelling for 3 hours each, and say if A has covered "d" miles, then B has covered the slack of 2490 - d.
![\bf \leftarrow \underset{A}{\stackrel{r}{\rule[0.22em]{8em}{0.25pt}}}dallas\underset{B}{\stackrel{r+80}{\rule[0.22em]{18em}{0.25pt}}}\to \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ plane~A&d&r&3\\ plane~B&2490-d&r+80&3 \end{array}](https://tex.z-dn.net/?f=%5Cbf%20%5Cleftarrow%20%5Cunderset%7BA%7D%7B%5Cstackrel%7Br%7D%7B%5Crule%5B0.22em%5D%7B8em%7D%7B0.25pt%7D%7D%7Ddallas%5Cunderset%7BB%7D%7B%5Cstackrel%7Br%2B80%7D%7B%5Crule%5B0.22em%5D%7B18em%7D%7B0.25pt%7D%7D%7D%5Cto%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20plane~A%26d%26r%263%5C%5C%20plane~B%262490-d%26r%2B80%263%20%5Cend%7Barray%7D)
Answer:
Below.
Step-by-step explanation:
a = 3, b = -5 and c = -12.
Answer:
62+62=2
122=2
Step-by-step explanation:
The left side
122
122
does not equal to the right side
2
2
, which means that the given statement is false.
False
