Solving for Angle A with Trigonometric Relations

We are given three conditions involving angles A, B, and C:

• Condition 1: $\sin(A) = \cos(B)$
• Condition 2: $\cos(A) = \sin(C)$
• Condition 3: $A + B + C = 90^\circ$

Deriving Relationships from Conditions 1 and 2

From Condition 1, $\sin(A) = \cos(B)$, we know that the sine of an angle is equal to the cosine of its complement. Therefore, we can write:

$ A + B = 90^\circ $

From Condition 2, $\cos(A) = \sin(C)$, similarly, the cosine of an angle is equal to the sine of its complement. Thus:

$ A + C = 90^\circ $

Substituting into Condition 3

Now we use Condition 3, $A + B + C = 90^\circ$. We can express B and C in terms of A using the relationships derived above:

• From $A + B = 90^\circ$, we get $B = 90^\circ - A$.
• From $A + C = 90^\circ$, we get $C = 90^\circ - A$.

Substitute these expressions for B and C into Condition 3:

$ A + (90^\circ - A) + (90^\circ - A) = 90^\circ $

Calculating the Value of Angle A

Simplify the equation:

$ A + 90^\circ - A + 90^\circ - A = 90^\circ $

Combine like terms:

$ 180^\circ - A = 90^\circ $

Solve for A:

$ A = 180^\circ - 90^\circ $

$ A = 90^\circ $

The value of angle A is $90^\circ$. Let's verify:

• If $A = 90^\circ$, then $\sin(A) = \sin(90^\circ) = 1$. For $\sin(A) = \cos(B)$ to be true, $\cos(B) = 1$, which means $B = 0^\circ$.
• If $A = 90^\circ$, then $\cos(A) = \cos(90^\circ) = 0$. For $\cos(A) = \sin(C)$ to be true, $\sin(C) = 0$, which means $C = 0^\circ$.
• Checking $A + B + C = 90^\circ$: $90^\circ + 0^\circ + 0^\circ = 90^\circ$. The conditions are satisfied.