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3D Trig Question (1 Viewer)

GaDaMIt

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Three tourists T1, T2 and T3 at ground level are observing a landmark L. T1 is due north of L, T3 is due east of L, and T2 is on the line of sight from T1 and T3 and between them. The angles of elevation to the top of L from T1, T2 and T3 are 25', 32' and 36' respectively.

a) show tan angle LT1T2 = cot 36 / cot 25.. already done this bit
b) use the sine rule in triangle LT1T2 to find, correct to the nearest minute, the bearing of T2 from L haven't done this bit..

please show working .. i take ages to figure out what people have done ..
 

GaDaMIt

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true north? theres not supposed to be a bearing in either case.. and no the answer is 13'41'. Man this question is annoying the hell out of me
 

SoulSearcher

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Ok, I think I have got it.

Ok for part B you need to know the angle of LT1T3, which is 32'41'35'
The other angle in the triangle LT3T1 is 57'18'25' (shown in diagram)
Now let the height of the landmark be h, so that LT2 = h cot 32 and LT3 = h cot 36
Let the angle LT2T3 be X
Therefore, using sine rule on the triangle LT2T3,
sin X / h cot 36 = sin 57'18'25' / h cot 32
Therefore
sin X = h cot 36 * sin 57'18'25' / h cot 32
sin X = cot 36 * sin 57'18'25' / cot 32
therefore angle X = 46'22'10.34'
since the angle sum of a triangle is 180, the angle T3LT2 (the angle made by the lines LT2 and LT3) is equal to 180 - ( 46'22'10.34' + 57'18'25' ) = 76'19'24.66'
but since we are looking for the angle between the lines LT1 and LT2 and the angle T1LT3 is 90, then the angle between the lines LT1 and LT2 is
90 - 76'19'24.66' = 13'40'35.34' = 13'41' to the nearest minute.
 
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