As shown in the figure, in the triangle ABC, ab = AC = 10, s triangle ABC = 30, find the length of BC

As shown in the figure, in the triangle ABC, ab = AC = 10, s triangle ABC = 30, find the length of BC

So s = 1 / 2 * 10sinx * 2 * 10cosx = 30, get sin2x = 0.6, then BC = 2 * 10cosx, the angle is not a normal angle, but I believe you can calculate the answer to about 19 with a calculator, and do it yourself

In the triangle ABC, if the angle ABC is 120 degrees and ab + BC is 30, the shortest AC is?
And sine theorem, cosine theorem

Let AB = x, then BC = 30-x. from the cosine theorem, we know (AC) ^ 2 = (AB) ^ 2 + (BC) ^ 2 * (AB) * (BC) * (COS ∠ ABC); substitute AB and CD to get (AC) ^ 2 = x ^ 2 + (30-x) ^ 2 * x * (30-x) * (cos120 °). Simplify (AC) ^ 2 = x ^ 2-30x + 900 = (X-15)

As shown in the figure, in △ ABC, AC = 50cm, BC = 40cm, ab = 30cm, point P moves from point a along the edge of AB to point B at a speed of 1cm / s, and the other point Q moves from point B along the edge of BC to point C at a speed of 2cm / s. After T seconds, the distance between P and Q is exactly 5T & nbsp; cm, so t

∵ AC = 50, BC = 40, ab = 30 ∵ ac2 = BC2 + AB2, so △ ABC is a right triangle, ∠ B = 90 ° Pb = (30-t), BQ = 2T, PQ = 5tpb2 + bq2 = pq2 (30-t) 2 + (2t) 2 = (5T) 2T = 15 A: t is 15 seconds

As shown in the figure, when △ abcab = 7, AC = 11, point m is the midpoint of BC, ad is the bisector of ∠ BAC, and MF ‖ ad, the length of FC is______ ．

As shown in the figure, ∵ ad is the bisector of ∠ BAC, ∵ CDDB = acab = 117, ∵ point m is the midpoint of BC, ∵ cm + dmcm − DM = 117, the solution is CMDM = 92. ∵ MF ∥ ad, ∵ CFFA = cmmd = 92. ∵ CF + FA = 11, ∵ CF = 9