Three θ \theta s

Geometry Level 5

Let A B C ABC be an isosceles triangle with A B = A C AB=AC and A = 2 θ \angle A= 2 \theta , θ 1 { \theta }_{ 1 } is value of θ \theta for which inradius of A B C \triangle ABC is maximum and θ 2 { \theta }_{ 2 } and θ 3 { \theta }_{ 3 } are solutions of cos θ = θ \cos \theta = \theta and sin θ = cos θ \sin \theta =\cos \theta is first quadrant respectively, then __________ \text{\_\_\_\_\_\_\_\_\_\_} .

θ 1 { \theta }_{ 1 } > θ 2 { \theta }_{ 2 } > θ 3 { \theta }_{ 3 } θ 1 { \theta }_{ 1 } < θ 2 { \theta }_{ 2 } < θ 3 { \theta }_{ 3 } None of above θ 1 { \theta }_{ 1 } > θ 2 { \theta }_{ 2 } < θ 1 { \theta }_{ 1 } θ 2 { \theta }_{ 2 } < θ 3 { \theta }_{ 3 } < θ 1 { \theta }_{ 1 }

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

0 solutions

No explanations have been posted yet. Check back later!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...