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# Find angle between line and plane in pyramid

Question:

P-ABCD is a pyramid. The base ABCD is a square. PA is perpendicular to plane ABCD. PA = square root of 8. AB = 4. Find the angle between PB and the plane PAC. Solution:

Connect DB. DB intersects AC at point O. Point O lies on plane ABCD and also lies on plane PCA. Connect PO. Because PA is perpendicular to plane ABCD and because BO lies in plane ABCD, so, BO is perpendicular to PA. Because ABCD is a square, AC and BD are two diagonals. So, AC = BD and AC, BD is perpendicular and bisect at point O. So, BO is perpendicular to AC.

Because PA and AC intersect at point A, so PA and AC are two intersected lines. Two intersected lines determine a plane. So, BO is perpendicular to plane PAC. PO is the projection of PB on the plane PAC. So, the angle between PB and plane PAC is the angle BPO. Now, we need to find the angle BPO.

Because PA is perpendicular to plane ABCD. So, PA is perpendicular to AB. So, PAB is a right triangle.

In right triangle PAB,
PB2 = PA2 + AB2
= (square root of 8)2 + 42
= 8 + 16 = 24
PB = square root of 24
= square root of (4 × 6)
= 2 (square root of 6)
In square ABCD,
AB = AD = 4, angle A = 90o
BD = 4 × square root of 2
BO = 2 × square root of 2
In right triangle POB,
sin (angle BPO) = OB/PB
= (2 × square root of 2) / (2 × square root of 6)
= square root of (2/6)
= square root of (1/3)
= square root of 0.3333
= 0.5773
angle BPO = arc sin 0.5773
= 35.26o

Therefore, the angle between PB and plane PCA is 35.26o. Please watch the video for more details.