Question

using co-function theorem (complementary angle theorem) to answer the question: $cos 15^{circ} = sin(?^{circ})$ note: just write the number in the blank place only. question 6 using co-function theorem (complementary angle theorem) to answer the question: $\tan(42^{circ}) = cot (?)^{circ}$ question 7 $37^{circ}4536 + 26^{circ}4946 =?$ you must give the answer using this format: 15-23-41, (no space), which means $15^{circ}2341$

Explanation:

First Question (cos 15° = sin(?°))

Step1: Recall Co - Function Theorem

The Co - Function Theorem states that for an acute angle \(\theta\), \(\cos\theta=\sin(90^{\circ}-\theta)\).

Step2: Apply the theorem to \(\cos15^{\circ}\)

Let \(\theta = 15^{\circ}\), then \(90^{\circ}-\theta=90^{\circ}- 15^{\circ}=75^{\circ}\). So \(\cos15^{\circ}=\sin(75^{\circ})\).

Step1: Recall Co - Function Theorem for tangent and cotangent

The Co - Function Theorem states that for an acute angle \(\theta\), \(\tan\theta=\cot(90^{\circ}-\theta)\).

Step2: Apply the theorem to \(\tan(42^{\circ})\)

Let \(\theta = 42^{\circ}\), then \(90^{\circ}-\theta = 90^{\circ}-42^{\circ}=48^{\circ}\). So \(\tan(42^{\circ})=\cot(48^{\circ})\).

Step1: Add the seconds

Add the seconds part: \(36'' + 46''=82''\). Since \(60'' = 1' \), we convert \(82''\) to minutes and seconds. \(82\div60 = 1\) with a remainder of \(22\). So we have \(1'\) and \(22''\), and we carry over \(1'\) to the minutes place.

Step2: Add the minutes

Add the minutes part along with the carried - over minute: \(45'+49'+1' = 95'\). Since \(60'=1^{\circ}\), we convert \(95'\) to degrees and minutes. \(95\div60 = 1\) with a remainder of \(35\). So we have \(1^{\circ}\) and \(35'\), and we carry over \(1^{\circ}\) to the degrees place.

Step3: Add the degrees

Add the degrees part along with the carried - over degree: \(37^{\circ}+26^{\circ}+1^{\circ}=64^{\circ}\).

Step4: Combine the results

Putting it all together, we have \(64^{\circ}35'22''\), which in the required format is \(64 - 35 - 22\).

Answer:

75

Second Question (tan(42°) = cot(?)°)