Angle C 90 ° Angle B 30 ° CD perpendicular to AB, BD = 7√3 find AB

1. Consider a triangle CDB: angle CDB = 90 degrees (since CD is perpendicular to AB), angle DBC (angle B) = 30 degrees, DB = 7√3 and CD are legs, CB is hypotenuse (since it lies opposite an angle equal to 90 degrees).
The CD leg lies opposite the DBC corner, then:
CD = CB / 2 (in a right-angled triangle opposite an angle of 30 degrees lies a leg, which is exactly 2 times less than the hypotenuse).
By the Pythagorean theorem:
CB = √ (CD ^ 2 + DB ^ 2).
Let’s substitute the expression CB / 2 instead of СD:
CB = √ ((CB / 2) ^ 2 + (7√3) ^ 2) = √ (CB ^ 2/4 + 147) = √ ((CB ^ 2 + 588) / 4);
CB ^ 2 = (CB ^ 2 + 588) / 4;
4СВ ^ 2 = СВ ^ 2 + 588 (in proportion);
3СВ ^ 2 = 588;
CB ^ 2 = 588/3;
CB ^ 2 = 196;
CB = √196;
CB = 14.
Then:
CD = 14/2 = 7.
2. Height is the geometric mean of the two segments of the hypotenuse formed by it, that is:
CD ^ 2 = AD * BD;
7√3AD = 49;
AD = 49 / 7√3;
AD = 7 / √3;
AD = 7√3 / 3.
3. AB = AD + BD;
AB = 7√3 / 3 + 7√3 = (7√3 + 21√3) / 3 = 28√3 / 3.
Answer: AB = 28√3 / 3.