# Thread: trigo 3D application

1. ## trigo 3D application

I have done (a)(b)(c) but fail to do (d). The book's answer of (d) is 67.2 degree.

Three points A, B and C on horizontal ground are such that AB = 10m, BC = 21m and $\displaystyle cos \angle ABC = \frac 3 5$ . The point P lies on BC and $\displaystyle \angle APB = 90$ degree. Calculate

(a) the length of AC
(b) the length of AP

A vertical pole VA of height 19m is placed at A. Calculate, to the nearest tenth of a degree, the acute angle between

(c) VB and the horizontal
(d) the plane VBC and the horizontal

Here's my working of (d) but my answer is wrong.

let VB = a , VC = b
$\displaystyle a^2 = 19^2 + 10^2 = 461$
$\displaystyle b^2 = 19^2 + 17^2 = 650$

consider triangle VBC,
$\displaystyle cos B = \frac {a^2 + 21^2 + b^2}{41a}= \frac{252}{41 \sqrt{461}}$
let VX be the height of triangle VBC and VX perpendicular to BC,
$\displaystyle VX = a\sin B = 20.57$
angle between VBC and the horizontal = $\displaystyle \angle VXA = sin ^{-1} \frac {19}{20.57} = 67.45$ degree (wrong)

2. Hello cakeboby
Originally Posted by cakeboby
I have done (a)(b)(c) but fail to do (d). The book's answer of (d) is 67.2 degree.

Three points A, B and C on horizontal ground are such that AB = 10m, BC = 21m and $\displaystyle cos \angle ABC = \frac 3 5$ . The point P lies on BC and $\displaystyle \angle APB = 90$ degree. Calculate

(a) the length of AC
(b) the length of AP

A vertical pole VA of height 19m is placed at A. Calculate, to the nearest tenth of a degree, the acute angle between

(c) VB and the horizontal
(d) the plane VBC and the horizontal

Here's my working of (d) but my answer is wrong.

let VB = a , VC = b
$\displaystyle a^2 = 19^2 + 10^2 = 461$
$\displaystyle b^2 = 19^2 + 17^2 = 650$

consider triangle VBC,
$\displaystyle cos B = \frac {a^2 + 21^2 + b^2}{41a}= \frac{252}{41 \sqrt{461}}$
let VX be the height of triangle VBC and VX perpendicular to BC,
$\displaystyle VX = a\sin B = 20.57$
angle between VBC and the horizontal = $\displaystyle \angle VXA = sin ^{-1} \frac {19}{20.57} = 67.45$ degree (wrong)

I'm not quite sure why you think you need the point $\displaystyle X$. The height of triangle $\displaystyle VBC$ is $\displaystyle VP$, since this line (like $\displaystyle AP$) is perpendicular to $\displaystyle BC$. (See attached diagram.)

So the angle you want is $\displaystyle \angle VPA = \tan^{-1}\left(\frac{VA}{AP}\right)=\tan^{-1}\left(\frac{19}{8}\right)=67.2^o$.

3. Originally Posted by Grandad
Hello cakebobyI'm not quite sure why you think you need the point $\displaystyle X$. The height of triangle $\displaystyle VBC$ is $\displaystyle VP$, since this line (like $\displaystyle AP$) is perpendicular to $\displaystyle BC$. (See attached diagram.)

So the angle you want is $\displaystyle \angle VPA = \tan^{-1}\left(\frac{VA}{AP}\right)=\tan^{-1}\left(\frac{19}{8}\right)=67.2^o$.

Thanks for the reply but $\displaystyle \tan ^{-1}\frac {19}{8} = 63.43$
I though about the same problem before posting the question but 63.43 didn't agree with the book's answer so I tried finding another method by creating the point X to find the angle. However I still got the wrong answer.

4. Hello cakeboby
Originally Posted by cakeboby
Thanks for the reply but $\displaystyle \tan ^{-1}\frac {19}{8} = 63.43$...
Not on my calculator.

5. Originally Posted by Grandad
Hello cakebobyNot on my calculator.

Oh, sorry! That's my mistake! Thanks!

### trigo application

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