#2015_4 Height of a street lamp?

Let AB = 9 = c , BC = 7 = a and AC = 8 = b .

M is the midpoint of the side AC at which lamp post MP of height h (say) is located Again,it is given that the lamp post subtends an angle 15° at B .

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 M is 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 , q and r = 7+2+3= $\boxed{12}$

By Stewart's Theorem we have $BM=\frac 1 2*\sqrt{2AB^2+2BC^2 - AC^2}=7$ .
In right angled triangle with BM and lamp height as legs and BM making an angle of 15 with hypotenuse ,
height = BM * Tan15.
Tan15=Tan(45 - 30)= $2 - \sqrt3.\ \ height =7*(2 - \sqrt3).\ \ 7+2+3=12$ .

M is the midpoint of AC so using apollonius theorem AB^2 + BC^2 = 2(CM^2 + BM^2) 7^2 + 9^2 = 2 ( 4^2 + BM^2) Solving this we get BM = 7. Then let the height of lamp be MP So in triangle BMP , angle PBM is 15 tan 15 = tan(45-30) tan45 - tan 30/1+tan45tan30 so MP /BM = tan 15 solving this we get MP = 7(2 - sqrt3) so 7=p , 2 = q and 3 = r p+q+r = 12

In triangle ABC, we get cosA=2/3 by cos rule.

So, we get BM=7 by cos rule.

Then we know that tan(15)=sin(15)/cos(15)=2sin(15)cos(15)/2cos^2(15).

This gives sin(30)/1+cos(30)=1/2+sqrt(3)=2-sqrt(3).

As tan(15)=x/7, where x is height of post, we get x=7[2-sqrt(3)].

Thus a+b+c=7+2+3=12.