Answer to Question #3815 in Trigonometry for tanmay
<br>a) 2Acos θ/2, along bisector <br>b) Acos θ/2, at 45 degree from one vector <br>c) Acos θ/2, along bisector
<br>d)Asin θ/2, along bisector
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
As you can see on the image,
two equal vectors AB and AC form the isosceles triangle ABC,
the resultant is the diagonal of the rhomb ABDC,
BC is the other diagonal.
In the rhomb the diagonals intersect in the middle and at right angle. Thus AO is the height and the middle line of the triangle ABC.
As ABC is isosceles triangle, AO is also the bisector of the angle BAC. So the angle BAO is θ/2.
AO = AB* cos (BAO) = a cos ( θ/2)
AD = 2 AO = 2a cos ( θ/2)
can u please give me procedure?