In triangle ABC, A.B.C is the length of the side opposite angle A.B.C, and S is the area of triangle ABC Know s = a Λ 2 - (B-C) Λ2. Find Tana

In triangle ABC, A.B.C is the length of the side opposite angle A.B.C, and S is the area of triangle ABC Know s = a Λ 2 - (B-C) Λ2. Find Tana

a ²- (b-c) ²= a ²- b ²+ 2bc-c ²= 2bc-2bccosA
S=1/2bcsinA
∴2bc-2bccosA=1/2bcsinA
4-4cosA=sinA
∵sinA ²+ cos ² A=16-32cosA+16cos ² A+cos ² A=1
∴17cos ² A-32cosA+15=0
‡ cosa = 1 (rounded off) or 15 / 17
∴sinA=8/17
∴tanA=8/15

In triangle ABC, the area of the triangle is (1 / 4) * (a * a + b * B-C * c), and the angle c is calculated

You should have learned sine theorem and cosine theorem?
Triangle area s = 1 / 2 A * b * sinc = 1 / 4 (a) ²+ b ²- c ²）
By the cosine theorem, 2A * b * COSC = a ²+ b ²- c ²
Compared with the two formulas
1/4 tanC=1/4
tanC=1
C=45°

It is known that the area of triangle ABC is s = 4 / 1 (b square + C square - a square), where a, B and C are the opposite sides of angles a, B and C respectively (1) Find the size of a; (2) If a = 2, B + C = 3, find the area of triangle ABC

First, explain to LZ that the representation of quarter is 1 / 4, not 4 / 1,
Let's start solving the problem
Cosine theorem
Party B + Party C - Party A
=2bc CosA
=1/2bcSinA*4(Cos A/Sin A)
=4S/tanA
From the known 4S = Party B + Party C - Party A
So tan a = 1
A=45°
Party B + Party C - Party A = 2BC cosa
So 2BC = 1 / cosa (square B + square c-square a) = 4S / cosa = 4 √ 2 s
(B + C) - a = 9-4 = 5
(B + C) square - a square = (b square + C square - a square) + 2BC = 4S + 4 √ 2S = 4 (1 + √ 2) s = 5
So s = 5 / 4 (√ 2-1)

In the triangle ABC, we know that the ratio of the degree of angle a, angle B and angle c is 1:2:3bc, which is equal to 4, and find the area of the triangle ABC

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In triangle ABC, given the area s = 1 / 4 (a Λ 2 + B Λ 2 + C Λ 2), what is the degree of angle c

Are you sure the question is not s = 1 / 4 (a ^ 2 + B ^ 2-C ^ 2) if so, the answer is 45 ° first C ^ 2 = a ^ 2 + B ^ 2-2abcosc into the original formula, and because s = 1 / 2absinc into C = 45 ° if the question is correct, I don't know

In triangle ABC, if its area is known to be s = 1 / 4 (a ^ 2 + B ^ 2-C ^ 2), what is the degree of angle c? To the detailed process, thank you!

A ^ 2 + B ^ 2-C ^ 2 = 2Ab * cos angle c (cosine theorem), so s = 1 / 2 (AB * cos angle c) because s = 1 / 2 (AB * sin angle c) (area formula), cos angle c = sin angle c, C = 45 ° in the triangle