SOLUTION: the base,N,of an antenna rests on horizontal ground. the angle of elecation of the top R, from a point H on the ground is 42(degrees) the angle of elevation of R from a second poi

Algebra ->  Trigonometry-basics -> SOLUTION: the base,N,of an antenna rests on horizontal ground. the angle of elecation of the top R, from a point H on the ground is 42(degrees) the angle of elevation of R from a second poi      Log On


   

Question 793000: the base,N,of an antenna rests on horizontal ground. the angle of elecation of the top R, from a point H on the ground is 42(degrees)
the angle of elevation of R from a second point, G, 6cm closer to N than H, is 53.2(degrees)
a)calculate
i)the length of RG
ii)the height RN, of the antenna
b)a third point c, lies on the ground 5.4m form N. the angle of depression of CfromR is x degrees
calculate to the nearest degree the vaule of x
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
It's all about tangents:
NH=NG%2B6m
RN%2FNG=tan%2853.2%5Eo%29=1.3367 (rounded)
RN%2FNH=RN%2F%28NG%2B6m%29=tan%2842%5Eo%29=0.9004 (rounded)
So %28NG%2B6m%29%2FNG=1.3367%2F0.9004 (approximately)
1%2B6m%2FNG=1.4846 (approximately)
6m%2FNG=0.4846
NG=6m%2F0.4846=highlight%2812.38m%29
NH=12.38m%2B6m=highlight%2818.38m%29
And since RN%2FNH=0.9004 (rounded) so RN=%2818.38m%29%2A0.9004=highlight%2816.55m%29

b) tan%28%28x%29%29=RN%2FNC=16.55m%2F%225.4+m%22=3.065 (rounded)
Since tan%2871.93%5Eo%29=3.065, highlight%28x=71.9%5Eo%29 (approximately)