Question
Show that in any triangle $ABC a = b \cos C + c \cos B$
✓
Answer
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
$4$ 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 $\angle C$ is obtuse in below figure.
From the right angled triangle $ADB$,
$BD \over AB = \cos B$
i.e. $BD = AB \cos B$
$= c \cos B$ 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 $ACB,$
$BC \over AB = \cos B$
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.
$a = b \cos C + c \cos B”$
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
$4$ 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 $\angle C$ is obtuse in below figure.
From the right angled triangle $ADB$,
$BD \over AB = \cos B$
i.e. $BD = AB \cos B$
$= c \cos B$ 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 $ACB,$
$BC \over AB = \cos B$
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.
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