In each case given below, find the value of angle A where `0^(@)leAle90^(@)`
(i) `sin(90^(@)-3A).cosec42^(@)=1`
(ii) `cos(90^(@)-A).sec77^(@)=1`

Answer

To solve the given problems, we will use the properties of complementary angles and trigonometric identities.

### Problem (i):
Given the equation:
\[ \sin(90^\circ - 3A) \cdot \csc(42^\circ) = 1 \]

**Step 1: Use the identity for sine of a complementary angle.**
Using the identity \(\sin(90^\circ - x) = \cos(x)\), we can rewrite the equation:
\[ \cos(3A) \cdot \csc(42^\circ) = 1 \]

**Step 2: Rewrite cosecant in terms of sine.**
Recall that \(\csc(x) = \frac{1}{\sin(x)}\), so we have:
\[ \cos(3A) \cdot \frac{1}{\sin(42^\circ)} = 1 \]

**Step 3: Rearrange the equation.**
Multiplying both sides by \(\sin(42^\circ)\):
\[ \cos(3A) = \sin(42^\circ) \]

**Step 4: Use the identity for cosine.**
Since \(\cos(3A) = \sin(90^\circ - 3A)\), we can also express this as:
\[ \cos(3A) = \sin(42^\circ) \]

**Step 5: Set up the equation.**
We can set up the equation:
\[ 3A = 90^\circ - 42^\circ \]
\[ 3A = 48^\circ \]

**Step 6: Solve for A.**
Dividing both sides by 3:
\[ A = \frac{48^\circ}{3} = 16^\circ \]

### Problem (ii):
Given the equation:
\[ \cos(90^\circ - A) \cdot \sec(77^\circ) = 1 \]

**Step 1: Use the identity for cosine of a complementary angle.**
Using the identity \(\cos(90^\circ - x) = \sin(x)\), we can rewrite the equation:
\[ \sin(A) \cdot \sec(77^\circ) = 1 \]

**Step 2: Rewrite secant in terms of cosine.**
Recall that \(\sec(x) = \frac{1}{\cos(x)}\), so we have:
\[ \sin(A) \cdot \frac{1}{\cos(77^\circ)} = 1 \]

**Step 3: Rearrange the equation.**
Multiplying both sides by \(\cos(77^\circ)\):
\[ \sin(A) = \cos(77^\circ) \]

**Step 4: Use the complementary angle identity.**
Since \(\cos(77^\circ) = \sin(90^\circ - 77^\circ)\):
\[ \sin(A) = \sin(13^\circ) \]

**Step 5: Set up the equation.**
Since both angles are acute, we can equate:
\[ A = 13^\circ \]

### Final Answers:
(i) \( A = 16^\circ \)
(ii) \( A = 13^\circ \)