Appendix 1 — Maths STD 12 Science — Question

Let p be the statement “ABC is any triangle” and q be the statement “
a = b cos C + c cos B”
Let ABC be a triangle. From A draw AD perpendicular to BC (BC produced if necessary).
As we know that any triangle has to be either acute or obtuse or right-angled, we can split p into three statements r, s and t, where
r : ABC is an acute-angled triangle with $\angle$C is acute.
s : ABC is an obtuse-angled triangle with $\angle$C is obtuse
t : ABC is a right-angled triangle with ∠C is a right angle.
Hence, we prove the theorem by three cases
case (i) When C is acute (Fig.
From the right-angled triangle ADB,
$\frac{B D}{A B}=\cos E$
$\mathrm{BD}=\mathrm{AB} \cos \mathrm{B}$
$=c \cos B$

from the right-angled triangle ADC 
$\frac{C D}{\Delta C}=\cos C$
CD = AC cos C
= b cos C
a = BD + CD
= c cos B + b cos C ...(1)
case(ii) When ∠ C is obtuse in below figure.

From the right angled triangle ADB,
$BD \over AB$ = cos B
i.e. BD = ABcos B
= c cos B From the right angled triangle ADC,
$\frac{C D}{A C}=\cos \angle A C D$
= cos (180^o - C) = - cos C i.e. CD = - AC cos C = - b cos C
Now a = BC = BD - CD
i.e. a = c cos B - ( - b cos C)
a = c cos B + b cos C ... (2)
Case (iii) When $\angle C$ is a right angle in the given figure.

From the right angled triangle ACB,
$BC \over AB$ = cosB
i.e. BC = AB cos B
a = c cos B,
and b cos C = b cos 900 = 0.
Thus, we may write a = 0 + c cos B
= b cos C + c cos B ... (3)
From (1), (2) and (3). We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r $\Rightarrow$ q is proved.
By case (ii), s $\Rightarrow$ q is proved.
By case (iii), t $\Rightarrow$ q is proved.
Hence, from the proof by cases, (r v s v t) $\Rightarrow$ q is proved, i.e., p $\Rightarrow$ q is proved.