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Introduction to Trigonometry

Long Answer Type

1. In ΔABC, right angled at B. AB = 24 cm, BC = 7 cm. Determine:
(i) sin A cos A,
(ii) sin C, cos C.

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2. In the given figure, find tan P - cot R.

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Short Answer Type

If $sinA space equals space 3 over 4 comma space$ calculate cos A and tan A.

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4. Given 15 cot A = 8, find sin A and sec A.

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Long Answer Type

Given $space secθ space equals space 13 over 12 comma space$ calculate all other trigonometric ratios.

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6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.

Let us consider two right angle triangles right angled at Q and S respectively.
Now,

cosA space equals space AS over AR

cosB space equals space BQ over BP

But cos A = cos B (given)
∴

AS over AR equals BQ over BP

rightwards double arrow

AS over BQ space equals space AR over BP

Let

AS over BQ equals AR over BP equals straight K

Let us consider two right angle triangles right angled at Q and S res

In ΔRAS,
Using Pythagoras theorem, we have

RS space equals space square root of AR squared minus AS squared end root space space space space space space space equals space square root of left parenthesis straight K times BP right parenthesis squared minus left parenthesis straight K. BQ right parenthesis squared end root space space space space space space space space equals space square root of left parenthesis straight K squared times BP squared minus straight K squared times BQ right parenthesis squared end root space space space space space space space equals space square root of straight K squared left parenthesis BP squared minus BQ squared right parenthesis end root space space space space space space equals space straight K square root of left parenthesis BP squared minus BQ squared right parenthesis end root

In APBQ,
Using Pythagoras theorem, we have

PQ space equals space square root of BP squared minus BQ squared end root

So,

RS over PQ equals fraction numerator straight K. square root of BP squared minus BQ squared end root over denominator square root of BP squared minus BQ squared end root end fraction equals straight K space space... left parenthesis ii right parenthesis

Comparing (i) and (ii), we get

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So, by using SSS similar condition

space increment RSA space tilde space increment PQB

∴

angle straight A space equals space angle straight B.

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If cot θ = $7 over 8 comma$ evaluate:
$fraction numerator left parenthesis 1 plus sinθ right parenthesis space left parenthesis 1 minus sinθ right parenthesis over denominator left parenthesis 1 plus cosθ right parenthesis space left parenthesis 1 minus cosθ right parenthesis end fraction$