A boat is crossing the river as shown. To the nearest tenth of a degree, find the angle at which the boat must head upriver in order to travel directly across the river. Diagram is not drawn to scale.
Just looking for some explaining on how to do this, not just the answer.
@skate-4-life270
ok so do u know the formula for trig?
No i dont
ok one sec let me look at something
k
\[a ^{?}+b ^{2}=c ^{2}\]
so c is the hypotenuse
so first find a^2 and b^2
do u know how to do that?
Would you plug 5 and 14 into that equation?
yes
\[4^{2}+14^{2}=c ^{2}\]
wait i typed that wrong
\[5^{2}+14^{2}=c ^{2}\]
Oh ok yah i think i can solve that! 5^2+14^2=c^2 25+196=c^2 221=c^2 I dont remember what to do with the ^2 on the c, i think it has something to do with square rooting
they want you to find the angle x You have to use trig, SOH CAH TOA in this case TOA , short for tangent x = opposite side/ adjacent side
yes now \[\sqrt{221}\]
c=14.9
yup thats the answer
is that one of the answers?
Only problem, is thats not the answer to any of the choices DX its either A> 70.3 North of west B> 19.7 North of west C. 19.7 south of west D. 70.3 South of West
A. 70.3 North of west B. 19.7 North of west C. 19.7 south of west D. 70.3 South of West
well shoot
@phi u mentioned something on here maybe he knows what he is talking about
im sorry thats as far into trig as ive gotten nevvy
\[ \tan x = \frac{5}{14} \] so \[ x = \tan^{-1}\frac{5}{14} \]
you should travel x degrees north of west
ok i will try that @phi
Join our real-time social learning platform and learn together with your friends!