maths is coooool (1 Viewer)

[noturningback]

okay i have a question that is literally driving me crazy
throwing textbooks across the room grrrrr

From the 04 Trial
B) A street light on a pole is 15 metres above the ground. A person, standing nearby, who is 1.8 metres tall casts a shadow from the light which is 5.4 metres long.

(ii) Calculate the distance of this person from the base of the light pole


Any help would be much appreciated, i can't find anything in my notes for this style of question. I think it has to do with similarity?

 

[lyounamu]

okay i have a question that is literally driving me crazy
throwing textbooks across the room grrrrr

From the 04 Trial
B) A street light on a pole is 15 metres above the ground. A person, standing nearby, who is 1.8 metres tall casts a shadow from the light which is 5.4 metres long.

(ii) Calculate the distance of this person from the base of the light pole


Any help would be much appreciated, i can't find anything in my notes for this style of question. I think it has to do with similarity?

1.

draw a triangle and another small one inside it.


then you will see that opposite side is 15cm and base is x.

small triangle is 1.8 and the base is 5.4


so you use

1.8/15 = 5.4/x (similar triangles = same ratio of sides)
x = 45

but distance of this person from the base of the light pole = x - 5.4 = 39.6

btw, nice thread title >.<

 

[lychnobity]

okay i have a question that is literally driving me crazy
throwing textbooks across the room grrrrr

From the 04 Trial
B) A street light on a pole is 15 metres above the ground. A person, standing nearby, who is 1.8 metres tall casts a shadow from the light which is 5.4 metres long.

(ii) Calculate the distance of this person from the base of the light pole


Any help would be much appreciated, i can't find anything in my notes for this style of question. I think it has to do with similarity?

You're right. Use similar triangles, from my diagram:

In triangles PAC & DBC:
angleC is common
anglePAC = angleDBC = 90
angleAPC = angleBDC (3rd angle in a triangle)
.'. triangle PAC is similar to triangle DBC

tan angleDCB = 1.8/5.4
ie angleDCB = 18

Similarly,
Tan anglePCA = 15/ac
tan 18 = 15/ac
AC = 15/tan 18 = 45

AC = AB + BC
AB = AC - BC
= 45 - 5.4 = 39.6m

 

[noturningback]

Thanks for the responses, i understood completely up until the x minus 5.4
I.e. i understand what you are doing there however I'm not entirely sure why it is used. Regardless, it solved my problem and I just need to memorize those steps.