In $ \triangle \mathrm{ABC}, \angle \mathrm{A}=\angle \mathrm{B}+\angle \mathrm{C} $ and $ \mathrm{AM} $ is an altitude. If $ \mathrm{AM}=\sqrt{12} $ and $ \mathrm{BC}=8 $, find BM.

Given:

In \( \triangle \mathrm{ABC}, \angle \mathrm{A}=\angle \mathrm{B}+\angle \mathrm{C} \) and \( \mathrm{AM} \) is an altitude.

\( \mathrm{AM}=\sqrt{12} \) and \( \mathrm{BC}=8 \).

To do:

We have to find BM.

Solution:

$\angle A+\angle B+\angle C=180^o$

$\angle A+\angle A=180^o$

$\angle A=90^o$

Let $BM=x$, this implies, $CM=8-x$

In $\triangle ABC$,

By Pythagoras theorem,

$BC^2 =AB^2+AC^2$

$(8)^2=AB^2+AC^2$

$64=AB^2+AC^2$.........(i)

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

$AB^2 =AM^2+BM^2$

$AB^2=(\sqrt{12})^2+x^2$

$AB^2=12+x^2$......(ii) 

In $\triangle AMC$,

By Pythagoras theorem,

$AC^2 =MC^2+AM^2$

$AC^2=(8-x)^2+(\sqrt{12})^2$

$AC^2=64+x^2-16x+12$......(iii)

From (i), (ii) and (iii),

$64=12+x^2+76+x^2-16x$

$2x^2+24-16x=0$

$x^2-8x+12=0$

$(x-6)(x-2)=0$

$x=6$ or $x=2$

Therefore,

$BM=2$ or $BM=6$