Sam's height is 1.8 meters.
The angle A'OB' = 112.8337°.
We assume the height of Sam, that is, AC or BD to be x meters.
From the given figure, we can say that the triangle A'CA is similar to the triangle A'OP, and the triangle B'CB is similar to B'OP.
Thus, for:
Triangles A'CA and A'OP, we can write:
CA/PO = AA'/PA',
or, x/6 = 2.1/(AP + 2.1),
Triangles B'DB and B'OP, we can write:
DB/PO = BB'/PB',
or, x/6 = 3.6/(BP + 3.6).
Thus, we get:
2.1/(AP + 2.1) = 3.6/(BP + 3.6),
or, 2.1BP + 7.56 = 3.6AP + 7.56,
or, 2.1BP = 3.6AP.
But, AP + BP = 13.3,
or, BP = 13.3 - AP.
Substituting this, we get:
2.1(13.3 - AP) = 3.6AP,
or, 27.93 - 2.1 AP = 3.6AP,
or, 2.1AP + 3.6AP = 27.93,
or, 5.7AP = 27.93,
or, AP = 27.93/5.7 = 4.9.
Substituting AP = 4.9 in x/6 = 2.1/(AP + 2.1), we get:
x/6 = 2.1(4.9 + 2.1),
or, x/6 = 2.1/7,
or, x = 2.1*6/7 = 1.8.
Thus, Sam's height = 1.8 meters.
Substituting AP = 4.9 in AP + BP = 13.3, we get:
4.9 + BP = 13.3,
or, BP = 13.3 - 4.9 = 8.4.
In triangle, A'OP,
tan ∠A'OP = A'P/PO,
or, ∠A'OP = tan ⁻¹(A'P/PO) = tan ⁻¹((A'A + AP)/PO) = tan ⁻¹((2.1 + 4.9)/6) = tan ⁻¹ (7/6) = 49.3987°.
In triangle B'OP,
tan ∠B'OP = B'P/PO,
or, ∠B'OP = tan ⁻¹(B'P/PO) = tan ⁻¹((B'B + BP)/PO) = tan ⁻¹((3.6 + 8.4)/6) = tan ⁻¹ (12/6) = tan ⁻¹ 2 = 63.4349°.
Now, ∠A'OB' = ∠A'OP + ∠B'OP = 49.3987° + 63.4349° = 112.8337°.
Thus, the angle A'OB' = 112.8337°.
Learn more about similar triangles at
brainly.com/question/28186893
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