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highschoolmom2010
you see a rock climber on a cliff at a 32 degree angle. your eve level is 6 feet from the ground and you stand 1000 feet from the base of the cliff. what is the approx. height of the rock climber from the ground?
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You know how to write sin for a triangle?
\[\sin \frac{ opp }{ hyp }\]
Yes try using this here we know sin30=1/2
where did you get that from
sin30 is a standard angle...It is good to know value of sin 0,30,45,60,90 otherwise you can always use calculator :P
Now can you proceed from here?
eh no not really
basically the begining this confused me bc i was given the adjacent side to the angle
Sorry I couldn't understand you...can you explain more
you use cos not sin if you are given adjecent. But better yet this problem requires tan :P
tan because you are given the adjecent. And wanna know the oppossite side.
so i need to use tan
Yup :D it is the simplest trig ID that has both opp and adj in it.
tan(30)=opp/adj=x/1000 tan(30)=x/1000 solve for x can you do that?
\[1000*\tan 30=\frac{ x }{ 1000 }*1000\] \[1000( \tan 30)=x\] but the angle is 32 degrees
1000*tan 32 =x is right so far :D what you do now is evaluate tan 32 in degree mode SOrry at typo
i must have look at that other guy who say sin 30 lol.
1000 tan (32) gave me a crazy answer
1000*(tan 32) =x ok so we got this far all we do is use our calculator making sure we in degree mode not radian mode tan32=0.62486935190932750978051082794943665831087020284270496 so we jsut plug that in 1000*.6248=624.8 ft sounds believeable. ooh plus 6 feet for the guys height at eye level 624.8+6
so rock climber is approx. 630.8' from the ground
if that is one of your answer choices :D use the one that is closest to our answer. oh and i would go with that lol.
Ohh sorry I made a mistake...was multitasking :(
i have no answer choices