Angle: The angles of elevation of the top of the tree was viewed by two boys as shown....

If you have gone through the steps of #18 you will see that statement above (derived by is incorrect.
That is precisely why I advocated the much simpler (more straightforward) approach of using the Sine Rule to (directly) calculate "h", followed by basic R-A Trig to get the boys' distances from the tree via the Tangents of the given angles but @chijioke appears to "enjoy(?)" making things difficult (for himself)! ?

(I wasn't about to waste any time ploughing through his (convoluted) algebraic manipulations to verify them; I will leave that in your very capable hands.
?)
 
Use pencil/paper .

Let
A = tan (48) ..... &....... B = tan(32).............Then.

32 * A * B/ ( A + B)...............divide numerator and denominator by (A * B) .......... I trust you know how to do that

= 32 * 1/[(1/A) + (1/B)] ......... I trust you can continue using the definition cot(x) = 1/tan(x)
Yes. I have seen how [math]\frac{32tan(48)\times tan(32)}{tan(32)+tan(48)}[/math] gave rise to [math]\frac{32}{cot48+cot32}[/math] So how will [math]\frac{32}{cot(48)+cot(32)}[/math] simplify further to give h = 12.79621962m or more precisely how will the denominator [math]cot(48)+cot(32)[/math] simplify ?
I am also asking; what did you think that made you divide the numerator and denominator by AB?
Your answers (for the height of the tree and the distance of each boy from it) now agree with mine (to the level of accuracy you have provided). :thumbup:
Thank you.
(And the diagram "provided" led to your mistakes in the first place!)
Yes. The mistake was because I did not look very well to see that [math]\rhd \text{ABC}[/math] was not an isosceles triangle (because the base angles were not equal)
may be true, but...

is not true, because [math]tan(32)\ne 0.624869351[/math] even to nine decimal places!

Any numerical value
(other than a few, like, for example, Sin 30° = 0.5 or Cos 0° = 1, which are exact values) that a calculator (or a table) provides for the "value" of a trigonometric ratio is simply an approximation!
True.
That is why you are (often) encouraged to work with surds in calculations involving trigonometric ratios, so that accuracy is preserved.
But it is not all trigonometric ratios that are in surd form. To the best of my knowledge, I think it is only trigonometric ratios of special angles that are in surd form. See surd.jpg
Do you have anything to say about it?
If you have gone through the steps of #18 you will see that statement above (derived by is incorrect.
So are you saying that [math]\frac{32tan(48)\times tan(32)}{tan(32)+tan(48)} \ne \frac{32}{cot48+cot32}[/math]
 
and now that you have a sketch of the problem you can (surely?) see what ∠ACB is and it is then a simple matter to calculate the length of AC (or BC) and thence h. (Using a certain Law with which I know you are familiar and Basic Right-Angled Triangle Trigonometry!)

The height of the tree is, of course,
h+1.5m and once you have gone thus far you should have no difficulty in finding the lengths of AT & BT.
Just incase I decide to use the Sine rule to solve it:
abc.jpg
[math]\frac{SinA}{a}=\frac{SinB}{b}[/math][math]a=\frac{bSinA}{SinB}[/math] [math]\rightarrow \frac{32Sin(48)}{Sin(100)}[/math][math]\frac{32 \times 0.743144825}{0.984807753}[/math][math]\frac{23.78063442}{0.984807753}= 24.14748903[/math]To obtain h
[math]sin(32 \degree) = \frac{h}{a}[/math][math]h= asin(32 \degree)[/math][math]h= 24.14748903 \times 0.529919264[/math][math]h= 12.79621962[/math]Complete height of tree = [math]12.79621962 + 1.5 = 14.29621962 \approx 14.3m[/math][math]tan(32 \degree) = \frac{h}{n}[/math][math]n=\frac{h}{tan(32 \degree)}= \frac{12.79621962}{0.624869351} = 20.47823213[/math][math]\therefore n \approx 20.5m[/math]This means one of the boy is standing a distance of 20.5m from the tree.
To obtain the distance of the other boy from the tree:
[math]32-n=32-20.47823213= 11.52176787 \approx 11.5m[/math]The other boy is standing at a distance of 11.5 m from the tree.
Thank you!
 
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