In an acute - ABC- if tan A - B - C - 1 and B - C - A - 2- find the angles A- B and C-

In ABC,a+B+C=π #160; #160; tan (A+B C)=1 #160; #160; tan (π 2C)=1π 2C=π4 C=3π8 #160; #160; sec(B+C A)=2(π 2A)=2π 2A=π3 A=π3 ...

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Standard X

Mathematics

Construction of Incircle

In an acute ...

Question

In an acute $Δ A B C$ , if $tan (A + B - C) = 1$ and $sec (B + C - A) = 2$ , find the angles $A, B$ and $C$ .

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Solution

In $△ A B C, a + B + C = π$
$tan (A + B - C) = 1$
$tan (π - 2 C) = 1 ⟹ π - 2 C = \frac{π}{4} ⟹ C = \frac{3 π}{8}$
$sec (B + C - A) = 2 ⟹ sec (π - 2 A) = 2 ⟹ π - 2 A = \frac{π}{3} ⟹ A = \frac{π}{3}$
$B = π - A - C = π - \frac{π}{3} - \frac{3 π}{8} = \frac{7 π}{24}$

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In an acute ΔABC, if tan(A+B−C)=1 and sec(B+C−A)=2, find the angles A,B and C.

In △ABC,a+B+C=π tan(A+B−C)=1 tan(π−2C)=1⟹π−2C=π4⟹C=3π8 sec(B+C−A)=2⟹sec(π−2A)=2⟹π−2A=π3⟹A=π3B=π−A−C=π−π3−3π8=7π24

In an acute $Δ A B C$ , if $tan (A + B - C) = 1$ and $sec (B + C - A) = 2$ , find the angles $A, B$ and $C$ .

In $△ A B C, a + B + C = π$
$tan (A + B - C) = 1$
$tan (π - 2 C) = 1 ⟹ π - 2 C = \frac{π}{4} ⟹ C = \frac{3 π}{8}$
$sec (B + C - A) = 2 ⟹ sec (π - 2 A) = 2 ⟹ π - 2 A = \frac{π}{3} ⟹ A = \frac{π}{3}$
$B = π - A - C = π - \frac{π}{3} - \frac{3 π}{8} = \frac{7 π}{24}$