The wording of this problem indicates that there was an illustration. Could you possibly share that illustration?
Working without an illustration:
If tan theta = 3, then tan theta = opp/adj = 3/1. This tells us that the opp side is 3 times as long as is the adj. side. Let x be the shorter side, i. e., let x represent the adjacent side; then y is the longer side and represents the opposite side.
Then y = 3x (the opp side is 3x the adj side in length).
Applying the Pyth. Thm. a^2 + b^2 = c^2,
x^2 + (3x)^2 = hyp^2 = (25 cm.)^2
So x^2 + 9x^2 = (625 cm^2)
10x^2 = 625 cm^2, or x^2 = (625 cm^2) / 10 = 62.5 cm^2
x = 7.91 cm. Therefore, y = 3(7.91) = 23.72 cm.
We were supposed to round off these answers to the nearest 10th cm.
Therefore, x = 7.9 cm and y = 23.7 cm
Would that be A, B, C or D?
Something that I’m not sure how to answer f(x) 16362
Sounds as tho' you have an isosceles triangle (a triangle with 2 equal sides). If this triangle is also a right triangle (with one 90-degree angle), then the side lengths MUST satisfy the Pythagorean Theorem.
Let's see whether they do.
8^2 + 8^2 = 11^2 ???
64 + 64 = 121? NO. This is not a right triangle.
If you really do have 2 sides that are both of length 8, and you really do have a right triangle, then:
8^2 + 8^2 = d^2, where d=hypotenuse. Then 64+64 = d^2, and
d = sqrt(128) = sqrt(8*16) = 4sqrt(8) = 4*2*sqrt(2) = 8sqrt(2) = 11.3.
11 is close to 11.3, but still, this triangle cannot really have 2 sides of length 8 and one side of length 11.
