Challenge yourself part 1

Geometry Level 4

In triangle A B C \triangle ABC , let D D be the mid point of B C BC . If A D B = 4 5 \angle ADB = 45^\circ and A C D = 3 0 \angle ACD = 30^\circ , find B A D \angle BAD in degrees.


The answer is 30.

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3 solutions

Nitish Pal
Sep 9, 2015

Either Change the Question to Angle BAC or the Answer is 15 degrees... Median Bisects the Angle at the Vertex opposite to the Side containing the Midpoint..

The question is right .

Jayesh Meena - 5 years, 9 months ago

no ,you are wrong ,nitish.

manu dude - 5 years, 9 months ago

I'm with u nitish. Answer is 15.

darshan parmar - 5 years, 8 months ago
Jose Lavariega
Oct 24, 2015

We are told that there is a triangle ABC , with D as the midpoint of side BC , and we are given some angles to help us. We will begin by stating that each of the 2 segments in which BC is divided by the midpoint, are of length 1. That is to say, both BD and DC are of length 1.

Using simple Geometry, we can find some angle lengths:

-Angle ADC has a measure of 135 degrees

-Angle DAC has a measure of 15 degrees

Now, our objective is to find the measure of side AD

Using the law of sines, and the identity

sin (a-b)= sin(a) cos(b) - cos(a) sin(b)

For the angle of 15, we get that its sine is

(root 6 - root 2)/4

we then apply the law of sines, to get that AD is sin 30 divided by the sin of 15

We already know that the sin of 15 is the expression above, and the sine of 30 is 1/2,

furthermore, we see that the side AD is (sqrt 6 + sqrt 2)/2

Now, we can apply the law of cosines to triangle BAD using side AD and BD, and the angle 45, which we already have. Remember that side BD is congruent to side DC, which we have stated are both of length 1.

Solving the law of cosines, we get that side AB is of length root 2

Now we use a law of sines to get the angle we are missing, which is 30

This question came in RMO 2005 .

Aniruddha Bagchi - 5 years ago
Divij Jain
Sep 19, 2015

Draw the specific triangle and we get ans. That is 30

You draw & find the answer. You had done it before.

Jayesh Meena - 5 years, 8 months ago

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Somehow I don`t think drawing the triangle is what's being asked here. I got the right answer by guestimating in my had, but there has to be a formulaic answer to this.

Ricardo Almeida - 5 years, 8 months ago

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