We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

In triangle ABC, angle ACB is 50 degrees, and angle CBA is 70 degrees. Let D be the foot of the perpendicular from A to BC, O the center of the circle circumscribed about triangle ABC, and E the other end of the diameter which goes through A. Find the angle DAE, in degrees.

codehtml127 Jun 29, 2018

#1**+2 **

Extend AD through to intersect the circle at F

Now.....since angle ACB = 50°...then minor arc AB has twice its measure =100°

And since angle CBA =70°....then angle BAD = BDA - CBA = 90 - 70 = 20° = angle BAF

And angle BAF = intercepts minor arc BF...so the measure of minor arc BF is twice this = 40°

But arc ABE =180° = minor arc AB + minor arc BF + minor arc FE

So

180 = 100 + 40 + minor arc FE

180 =140 + minor arc FE

40 = minor arc FE

But angle FAE intercepts minor arc FE, so it has 1/2 of its measure = 20°

But angle FAE = angle DAE...so DAE = 20°

CPhill Jun 29, 2018

#2**+2 **

Also....

There are 180° in triangle ACD, so..

m∠DAC = 180° - 90° - 50°

m∠DAC = 40°

An inscribed angle is half the measure of its intercepted arc, so....

(1/2)m∠AOC = m∠ABC

(1/2)m∠AOC = 70°

m∠AOC = 140°

OA and OC are radii of circle O , so they are the same length, and triangle AOC is isosceles. So...

m∠OAC = m∠OCA

m∠OAC + m∠OCA + 140° = 180°

m∠OAC + m∠OAC + 140° = 180°

2m∠OAC + 140° = 180°

2m∠OAC = 40°

m∠OAC = 20°

m∠DAE = m∠DAC - m∠OAC

m∠DAE = 40° - 20°

m∠DAE = 20°

hectictar Jun 29, 2018