Question 29
ABC is right angled triangle at C. Let BC = a, CA = b and AB = c and let p be the length of perpendicular from C on AB, then cp is equal to
Solution
Given that BC = a, AB = c, AC = b, CD = p
CD is perpendicular to AB & <ACB = 90 degrees.
Let <DBC = <ABC = $$\theta\ $$
$$Sin\ \theta\ =\dfrac{\ CD}{CB}=\dfrac{\ AC}{AB}$$
= $$\dfrac{\ p}{a}=\dfrac{\ b}{c}$$
Therefore, pc = ab.
