(i) My diagram looks like as below
![my computer]()
(ii) In Delta ADC, after converting all distances to meters.
tan alpha=(CD)/20000
=>CD=2.0xx10^4tan alpha " meter"
(iii) We make use of cosine rule.
For a triangle with sides a,b and c and angles A, B and C the cosine rule states:
a^2 = b^2 + c^2 - 2bc cos A
Applying in DeltaABC where angle A=120^@
BC^2 = (2.0xx10^4)^2 + (1.0xx10^4)^2 - 2xx2.0xx10^4xx1.0xx10^4 cos 120^@
We know that cos120^@=-1/2
=>BC^2 = 4.0xx10^8 + 1.0xx10^8 + 2.0xx10^8
=>BC^2 = 7.0xx10^8
=>BC = sqrt(7.0xx10^8)
=>BC = sqrt7xx10^4 " meter"
(iv) In DeltaBDC
tan25^@15'=(CD)/(BC)
Making use of values found in (ii) and (iii)
tan25^@15'=(2.0xx10^4tan alpha)/(sqrt7xx10^4)
Solving for alpha
tan alpha=sqrt7 /2tan25^@15'
alpha=tan^-1(0.6239)
alphaapprox32^@, correct to nearest degree.
