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Question: If i) \[\tan 2A = \cot (A - {18^ \circ })\], where \[2A\]and \[A - {18^ \circ }\]are acute angles. F...

If i) tan⁡2A=cot⁡(A−18∘)\tan 2A = \cot (A - {18^ \circ }), where 2A2Aand A−18∘A - {18^ \circ }are acute angles. Find ∠A\angle A
ii) sec⁡2A=csc⁡(A−27∘)\sec 2A = \csc (A - {27^ \circ }), where 2A2Ais an acute angle. Find the measure of ∠A\angle A.

Explanation

Solution

Hint : Here the question is related to the trigonometry. In trigonometry we have complementary angles for the ratios. Using that concept and the simple arithmetic operations we determine the required solution for the given question.

Complete step-by-step answer :
In the trigonometry we have six trigonometry ratios namely sine , cosine, tangent, cosecant, secant and cotangent. These are abbreviated as sin, cos, tan, csc, sec and cot. The 3 trigonometry ratios are reciprocal of the other trigonometry ratios. Here cosine is the reciprocal of the sine. The secant is the reciprocal of the cosine. The cotangent is the reciprocal of the tangent.
The trigonometric ratios for the complementary angles is given by
sin⁡(90−A)=cos⁡A\sin (90 - A) = \cos A
cos⁡(90−A)=sin⁡A\cos (90 - A) = \sin A
tan⁡(90−A)=cot⁡A\tan (90 - A) = \cot A
cot⁡(90−A)=tan⁡A\cot (90 - A) = \tan A
sec⁡(90−A)=csc⁡A\sec (90 - A) = \csc A
csc⁡(90−A)=sec⁡A\csc (90 - A) = \sec A
Now we will consider the given question
i) tan⁡2A=cot⁡(A−18∘)\tan 2A = \cot (A - {18^ \circ })
By using the trigonometric ratios for complementary angles, i.e., cot⁡(90−A)=tan⁡A\cot (90 - A) = \tan A. The above inequality is written as
⇒cot⁡(90∘−2A)=cot⁡(A−18∘)\Rightarrow \cot ({90^ \circ } - 2A) = \cot (A - {18^ \circ })
Since both sides the cotangent trigonometric ratios are present. We can cancel it and it is written as
⇒90∘−2A=A−18∘\Rightarrow {90^ \circ } - 2A = A - {18^ \circ }
Take the A terms one side and the angles on other side, we have
⇒90∘+18∘=2A+A\Rightarrow {90^ \circ } + {18^ \circ } = 2A + A
On adding the terms we have
⇒3A=108∘\Rightarrow 3A = {108^ \circ }
On dividing by 3 we have
⇒A=36∘\Rightarrow A = {36^ \circ }
Hence we have determined the angle A.
So, the correct answer is “⇒A=36∘ \Rightarrow A = {36^ \circ }”.

iI.) sec⁡2A=csc⁡(A−27∘)\sec 2A = \csc (A - {27^ \circ })
By using the trigonometric ratios for complementary angles, i.e., csc⁡(90−A)=sec⁡A\csc (90 - A) = \sec A. The above inequality is written as
⇒csc⁡(90∘−2A)=csc⁡(A−27∘)\Rightarrow \csc ({90^ \circ } - 2A) = \csc (A - {27^ \circ })
Since both sides the cosecant trigonometric ratios are present. We can cancel it and it is written as
⇒90∘−2A=A−27∘\Rightarrow {90^ \circ } - 2A = A - {27^ \circ }
Take the A terms one side and the angles on other side, we have
⇒90∘+27∘=2A+A\Rightarrow {90^ \circ } + {27^ \circ } = 2A + A
On adding the terms we have
⇒3A=117∘\Rightarrow 3A = {117^ \circ }
On dividing by 3 we have
⇒A=39∘\Rightarrow A = {39^ \circ }
Hence we have determined the angle A.
So, the correct answer is “⇒A=39∘ \Rightarrow A = {39^ \circ }”.

Note : Students may not get confused by seeing the question. As it involves doubles we need not substitute the formula of the doubles. By using the trigonometric ratios for the complementary angles it is a very easy way to solve the question. Furthermore, simple arithmetic operations are used.