In the triangle ABC, ad is the height on the BC side, tanb = cos angle DAC If sinc = 12 / 13, BC = 12, find the length of AD

Tanb = cos angle DAC
Ad / BD = ad / AC
So BD = AC
sinC=AD/AC=12/13
Let ad = 12x, AC = 13X, BD = AC = 13X
DC2+AD2=AC2
DC2+(12X)2=(13X)2
DC2 = 25x2, so DC = 5x
BC=BD+DC=13X+5X=18X=12
So x = 2 / 3
AD=12X=8

In the right triangle ABC, AC = BC = ad, angle DAC = 30 degrees, it is proved that BD = CD

It is proved that if BD = CD, then the angle DCB = angle DBC. From the known conditions, we can infer that the angle DCB = 15 degrees, so the angle DBC = 15 degrees

Given that the three sides of the triangle are a = 5 / 4, B = 1, C = 2 / 3, judge whether the triangle ABC is a right triangle

The longest side is a
1^2+(2/3)^2≠(5/4)^2
Triangle ABC is not a right triangle

Given the triangle ABC, a (2,3), B (- 2,4), C (- 1, - 9), it is proved that triangle ABC is a right triangle

Given the triangle ABC, a (2,3), B (- 2,4), C (- 1, - 9), it is proved that triangle ABC is a right triangle
The slope of line AB is (4-3) / (- 2-2) = - 1 / 4
The slope of the line AC is (3 + 9) / (2 + 1) = 4
The slopes of the two lines are opposite and reciprocal to each other. It can be seen that AB is perpendicular to AC,
So the triangle ABC is a right triangle

If the square of ABC + 6b is a right triangle with the square length of ABC + 6B, it is necessary to judge whether the square of the triangle ABC + 6b is a right triangle + 6B

In order to make the equation true: a = 3, B = 4, C = 55? = 3? + 4? Conforms to the Pythagorean theorem

It is known that the length of the three sides of △ ABC is a, B, C, satisfying the following conditions: A2 + B2 + c2-6a-8b-10c + 50 = 0, try to judge the shape of △ ABC

∵a2+b2+c2-6a-8b-10c+50=0，
∴a2-6a+9+b2-8b+16+c2-10c+25=0，
That is (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
∵32+42=52，
The △ ABC is a right triangle

If a, B and C are the three sides of △ ABC and A2 + B2 + C2 + 50 = 6A + 8b + 10C, judge the shape of the triangle

According to the known conditions, the original formula can be transformed into (A-3) 2 + (B-4) 2 + (C-5) 2 = 0,
∴a=3，b=4，c=5，
Then the triangle is a right triangle

If the three sides of the triangle ABC satisfy a + B + C + 50 = 6A + 8b + 10C

A square + b square + C square + 50 = 6A + 8b + 10C
Square + 0C - 8B
(a square - 6A + 9) + (b square - 8b + 16) + (C square - 10C + 25) = 0
(A-3) square + (B-4) square + (C-5) square = 0
So a = 3, B = 4, C = 5
3*3+4*4=5*5
So the triangle is a right triangle
Happy study

It is known that the three sides a, B and C of △ ABC satisfy a + B = 8, ab = 4, C = 2, radical 14. It is proved that: △ ABC is a right triangle

a^2+b^2=(a+b)^2-2ab=56
So a ^ 2 + B ^ 2 = C ^ 2;
So the right triangle of delta ABC

In △ ABC, the length of three sides is a = n-1, B = 2n, C = n + 1 (n > 1). Try to judge whether the triangle is a right triangle It is known that in △ ABC, the length of the three sides is a = n-1, B = 2n, C = n + 1 (n > 1). Try to judge whether the triangle is a right triangle. If so, please indicate which side is right angle to It is known that in △ ABC, the lengths of three sides are a = n square - 1, B = 2n, C = n square + 1 (n > 1). Try to judge whether the triangle is a right triangle. If so, please point out which side is the right angle

Because a + C = B, so ABC is not a triangle, is the title wrong

In the triangle ABC, ad is the height on the BC side, tanb = cos angle DAC If sinc = 12 / 13, BC = 12, find the length of AD