) In the figure, find the following:
i) sin50° ii)sin30° iii) cos60°iv) tan40° v)sec50° vi) cosec30°vii)cos30°.sin60° viii) sin40°.cosec60°ix) cot60°.sin50° x) sec60°.cot30
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Answer :-
Let,
- Perpendicular = P
- Base = B
- Hypotenuse = H
now, in ∆ABD, we have,
- ∠ABD = 30°
- ∠BAD = 180° - (90° + 30°) = 60°
so,
→ sin 30° = P/H = AD/AB = cos 60°
→ cosec 30° = H/P = AB/AD = sec 60°
→ sin 60° = P/H = BD/AB = cos 30°
→ cosec 60° = H/P = AB/BD = sec 30°
→ tan 30° = P/B = AD/BD = cot 60°
→ tan 60° = BD/AD = cot 30°
and, in ∆ADC, we have,
- ∠ACD = 50°
- ∠DAC = 180° - (90° + 50°) = 40°
so,
→ sin 40° = P/H = CD/AC = cos 50°
→ cosec 40° = H/P = AC/CD = sec 50°
→ sin 50° = P/H = AD/AC = cos 40°
→ cosec 50° = H/P = AC/AD = sec 40°
→ tan 40° = P/B = CD/AD = cot 50°
→ tan 50° = AD/CD = cot 40°
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