
In an acute angled ABC, if $ \tan (A + B - C) = 1 $ and $ \sec (B + C - A) = 2 $ , find the value of A, B and C.
Answer
593.1k+ views
Hint: By using the trigonometric table, we will form equations and further simplify the equations A, B and C values can be calculated. In RHS, we can replace 1 as $ tan {45^o} $. From there we will find equations in the form of A, B, and C. On solving these equations, we will get the value of A,B and C.
Complete step-by-step answer:
According to the question, we have
$ \Rightarrow \tan (A + B + C) = 1 $ and $ \sec (B + C - A) = 2 $
$ \Rightarrow \tan (A + B + C) = \tan {45^o}[\because \tan {45^o} = 1] $
and $ \sec (B + C - A) = \sec {60^o}[\because \sec {60^o} = 2] $
$ \Rightarrow A + B - C = {45^o} \to (1) $
$ B + C - A = {60^o} \to (2) $
Adding equation (1) and (2), we get
$
\Rightarrow (A + B - C) + (B + C - A) = {45^o} + {60^o} \\
\Rightarrow 2B = {105^o} \\
\Rightarrow B = 52{\dfrac{1}{2}^o} \\
$
Now as per the calculation above, putting $ B = 52{\dfrac{1}{2}^o} $ in equation (2), we get
$
\Rightarrow 52{\dfrac{1}{2}^o} + C - A = 60 \\
\Rightarrow C - A = 7{\dfrac{1}{2}^o} \to (3) \\
$
As we know, the sum of angles in a triangle is $ {180^o} $ .Therefore, we get
$
\Rightarrow A + B + C = {180^o} \\
\Rightarrow A + C = {180^o} - 52{\dfrac{1}{2}^o}[\because B = 52{\dfrac{1}{2}^o}] \\
\Rightarrow A + C = 127{\dfrac{1}{2}^o} \to (4) \\
$
Adding equation (3) and (4), we get
$
\Rightarrow 2C = {135^o} \\
\Rightarrow C = 67{\dfrac{1}{2}^o} \\
$
Substituting the obtained value in equation (3), we get
$
\Rightarrow 67{\dfrac{1}{2}^o} - A = 7{\dfrac{1}{2}^o} \\
\Rightarrow A = {60^o} \\
$
Hence, $ A = {60^o},B = 52{\dfrac{1}{2}^o},C = 67{\dfrac{1}{2}^o} $ .
Note: For this type of question, read the question carefully. Make equations separately and relate them to each other. You will find it useful. Remember basic trigonometric identities and values.
Complete step-by-step answer:
According to the question, we have
$ \Rightarrow \tan (A + B + C) = 1 $ and $ \sec (B + C - A) = 2 $
$ \Rightarrow \tan (A + B + C) = \tan {45^o}[\because \tan {45^o} = 1] $
and $ \sec (B + C - A) = \sec {60^o}[\because \sec {60^o} = 2] $
$ \Rightarrow A + B - C = {45^o} \to (1) $
$ B + C - A = {60^o} \to (2) $
Adding equation (1) and (2), we get
$
\Rightarrow (A + B - C) + (B + C - A) = {45^o} + {60^o} \\
\Rightarrow 2B = {105^o} \\
\Rightarrow B = 52{\dfrac{1}{2}^o} \\
$
Now as per the calculation above, putting $ B = 52{\dfrac{1}{2}^o} $ in equation (2), we get
$
\Rightarrow 52{\dfrac{1}{2}^o} + C - A = 60 \\
\Rightarrow C - A = 7{\dfrac{1}{2}^o} \to (3) \\
$
As we know, the sum of angles in a triangle is $ {180^o} $ .Therefore, we get
$
\Rightarrow A + B + C = {180^o} \\
\Rightarrow A + C = {180^o} - 52{\dfrac{1}{2}^o}[\because B = 52{\dfrac{1}{2}^o}] \\
\Rightarrow A + C = 127{\dfrac{1}{2}^o} \to (4) \\
$
Adding equation (3) and (4), we get
$
\Rightarrow 2C = {135^o} \\
\Rightarrow C = 67{\dfrac{1}{2}^o} \\
$
Substituting the obtained value in equation (3), we get
$
\Rightarrow 67{\dfrac{1}{2}^o} - A = 7{\dfrac{1}{2}^o} \\
\Rightarrow A = {60^o} \\
$
Hence, $ A = {60^o},B = 52{\dfrac{1}{2}^o},C = 67{\dfrac{1}{2}^o} $ .
Note: For this type of question, read the question carefully. Make equations separately and relate them to each other. You will find it useful. Remember basic trigonometric identities and values.
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