Question

n a triangle ABC if AC=BC=4cm , AC is proved to E and BC is produced to D .if angle A =40 degree, then find angle DCE

Answer 1

Answer:

Angle DCE = 100°

Step-by-step explanation:

Draw perpendicular CH such that angle AHC = BHC = 90°

A=45°, H=90° and angle ACH =50° because sum of angles in a triangle is 180°

Angle ACH= angle HCB ( since AC=BC) Thus angle ACB =100°

Since angle ACB and angle DCE are opposite and vertex angles they are equal

Therefore angle DCE = 100°

Answer 2

Step-by-step explanation:

Let the line segment A(5, -6) and B(-1, -4) is divided at point P(0, y) by y-axis in ratio m:n

:. x = \frac{mx2+nx1}{m+n

and y = \frac{my2+ny1}{m+n

Here, (x, y) = (0, y); (x1, y1) = (5, -6) and (x2, y2) = (-1, -4)

So , 0 = \frac{m(-1)+n(5)}{m+n}

=> 0 = -m + 5n

=> m= 5n

=> \frac{m}{n}

= \frac{5}{1}

=> m:n = 5:1

Hence, the ratio is 5:1 and the division is internal.Now,

y = \frac{my2+ny1}{m+n}

=> y = \frac{5(-4)+1(-6)}{5+1}

=> y = \frac{-20-6}{6}

=> y = \frac{-26}{6}

=> y = \frac{-13}{3}

Hence, the coordinates of the point of division is (0, -13/3).