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)

Thanks in advance!