The answer is 12.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

Let

AB = 9 = c,BC = 7 = aandAC = 8 = b.Mis the midpoint of the sideACat which lamp postMPof heighth(say) is located Again,it is given that the lamp post subtends an angle15°atB.Applying Cosine formula in $\Delta ABC$ , we have

$cosC=\frac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } =\frac { 49+64-81 }{ 2\times 7\times 8 } =\frac { 2 }{ 7 } \quad \quad \quad \quad --------(1)$

Similarly using cosine formula in $\Delta BMC$ , we get

${ BM }^{ 2 }={ BC }^{ 2 }+{ CM }^{ 2 }-2BC\times CMcosC$

Here $CM=\frac { 1 }{ 2 } CA=\quad 4$ , since

Mis the midpoint of AC.Therefore, using (1), we get

${ BM }^{ 2 }=49+16-2\times 7\times 4\times \frac { 2 }{ 7 } =49$

$\Rightarrow { BM }=\quad 7$

Thus, from $\Delta BMP$ right angled at

M, we have$tan15°=\frac { PM }{ BM } =\frac { h }{ 7 }$

$\Rightarrow 2-\sqrt { 3 } =\frac { h }{ 7 }$

$\Rightarrow h=7(2-\sqrt { 3 } )\quad m$

$\Rightarrow p=7\quad ,q=2\quad and\quad r=3$

Sum of

p,qandr=7+2+3= $\boxed{12}$