Question

Please solve this question. I Will mark your ans as brainliest answer. If you give irrelevant answers then I will report his/her answer.​

Answer 1

Answer:

In the figure we have regular pyramid.

All the edges have length l cm.

So, each face of the pyramid is an equilateral triangle.

From vertex D drop perpendicular to meet opposite face ABC at point G, which is the centre of the triangle ABC.

Here, G is also centroid of triangle ABC as it an equailateral triangle.

Clearly, AG=23AM

In triangle AMB,AM=ABsin60∘=3–√2lcm

∴AG=23×AM=23×3–√2l=123–√lcm

In triangle AGD, using Pythagoras, we get

AD2=AG2+GD2

∴l2=13l2+GD2

∴GD=23−−√1cm

For angle between any edge and face not containing that edge, consider edge AD and face ABC.

Angle between them is α,which is angle between AD and AM.

In triangle AGD, cosα=AGAD=13√ll=13–√

Angle between two faces ABC and BCD is the angle between AM and DM, which is β.

In triangle DGM,cosβ=GMDM=13