Given sin(∠QOP) = cos(∠) and cos(∠ROQ) = sin(∠), triangles QOP and ROQ are congruent.
In the given figure, we have two right triangles: QOP and ROQ. We are given that sin ∠QOP = cos ∠ and cos∠ROQ = sin ∠. This means that the two triangles are complementary.
Complementary triangles are two right triangles that share a hypotenuse and have acute angles that add up to 90 degrees. In other words, if we have two complementary right triangles, we can label them so that the following is true:
- One triangle has angles ∠A, ∠B, and ∠C.
- The other triangle has angles ∠C, ∠D, and ∠E.
- ∠A + ∠D = 90 degrees.
- ∠B + ∠E = 90 degrees.
In the figure given, we can label the triangles as follows:
- Triangle QOP has angles ∠QOP, ∠POQ, and ∠QPO.
- Triangle ROQ has angles ∠ROQ, ∠ORQ, and ∠QOR.
We are given that sin ∠QOP = cos ∠ and cos∠ROQ = sin ∠. This means that:
- sin ∠QOP = cos ∠OQR
- cos ∠ROQ = sin ∠POQ
Since triangles QOP and ROQ are complementary, we know that ∠QOP + ∠ROQ = 90 degrees. This means that ∠OQR = 90 degrees - ∠ROQ and ∠POQ = 90 degrees - ∠QOP.
Substituting these values into the equations above, we get:
- sin ∠QOP = cos (90 degrees - ∠ROQ)
- cos ∠ROQ = sin (90 degrees - ∠QOP)
Using the trigonometric identity that sin (90 degrees - ∠θ) = cos ∠θ, we can simplify these equations to:
- sin ∠QOP = sin ∠ROQ
- cos ∠ROQ = cos ∠QOP
This means that the two triangles, QOP and ROQ, have the same sine and cosine values for their corresponding angles. This is because the two triangles are congruent.
In conclusion, since sin ∠QOP = cos ∠ and cos∠ROQ = sin ∠, we can say that triangles QOP and ROQ are congruent.
Question:
In this figure, sin ∠QOP = cos ∠____
and cos∠ROQ = sin ∠___