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In △ ABC, if A2 + B2 = 25, A2-B2 = 7, C = 5, then the height of the largest edge is______ ．
According to the inverse theorem of Pythagorean theorem, a triangle is a right triangle, C is a hypotenuse, and the height of C is h. according to the area formula s = 12ab = 12CH, ∧ H = 125, we fill in 125
If the middle angle A-B of triangle ABC is 10 ° and the double angle C-3 is 25 ° then the angle a =?
There is a hidden condition, △ ABC has a + B + C = 180 degree
Then, according to the known conditions, a = B + 10 degree
Then 2B + C = 170 degree
2c-3b=25°
Then 7C = 560 degree
c=80°
Then B = 45 degree
That is: a = 55 degree
In △ ABC, if A2 + B2 = 25, A2-B2 = 7, C = 5, then the height of the largest edge is______ ．
According to the inverse theorem of Pythagorean theorem, a triangle is a right triangle, C is a hypotenuse, and the height of C is h. according to the area formula s = 12ab = 12CH, ∧ H = 125, we fill in 125
It is known that in the triangle ABC, the angles a, B, C and their corresponding sides are a, B, C and a = 7, B = 5, C = 3,
1 find the maximum angle of internal angle in triangle ABC, 2 find the area of triangle ABC
1. A has the largest corner, that is, a has the largest angle
cosA=(b²+c²-a²)/2bc
=(25+9-49)/2*5*3
=-1/2
∴A=120°
2、SΔ=(1/2)bcsinA
=(1/2)5*3sin60°
=5√3
In the triangle ABC, angle a = 120 degrees, a = 7, B + C = 8, find B, C and angle B
According to the cosine theorem: A ^ 2 = B ^ 2 + C ^ 2-2bccosa = 7 ^ 2 B ^ 2 + C ^ 2 + BC = 49 -- - (1) (COSA = cos120 ° = - 1 / 2) B + C = 8 (B + C) ^ 2 = 64b ^ 2 + C ^ 2 + 2BC = 64 -- - (2) (2) - (1): BC = 15 -- - (3) B (8-b) = 158b-b ^ 2 = 15b ^ 2-8b + 15 = 0 (B-3) (B-5) = 0b1 = 3, B2 = 5
In the triangle ABC, the opposite sides of the angles a, B and C are a, B and C, respectively. Tan C = 3 times root 7, find COSC
But the answer is one in eight
It's easy
Let me teach you a formula: (COSC) ^ 2 = 1 / [1 + (Tanc) ^ 2]
The formula has been used for a long time, so I made a mistake
So this problem can be directly set formula
(cosC)^2=1/(1+63)=1/64,
From the Tanc > 0, 0
In the triangle ABC, the opposite sides of angle a, angle B and angle c are ABC respectively, and Tanc is equal to three times of the root seven to find COSC
sin^2(C)+cos^2(C)=1
tan^2(C)+1=1/[cos^2(C)]
cos^2(C)=1/[tan^2(C)+1]
cosC=(+/-)1/8
Plus or minus one eighth
Ask for bonus points!
In the triangle ABC, the opposite sides of the angle a, B and C are respectively ABC, Tanc = 3 √ 7 and COSC
tanC=sinC/cos²C
=√（1-cos²C）/cos²C
=3√7
cosC=±1/8
∵tanC＞0
∴cosC=1/8
In the triangle ABC, the opposite sides of the angle ABC are ABC, and Tanc is equal to 3 √ 7?
Just a moment
In △ ABC, the opposite sides of a, B and C are a, B and C respectively, and 2b = a + C, then the value range of B is___ ．
∵ 2B = a + C, that is, B = a + C2, ∵ CoSb = A2 + c2-b22ac = A2 + C2 - (a + C) 242ac = 3 (A2 + C2) - 2ac8ac ≥ 4ac8ac = 12, then the range of B is (0, π 3)]
RELATED INFORMATIONS
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