Question

from the given figure find the following ratios tan 50 sin 50 cos60 cos 30 cos 50 cos 60​

Answer 1

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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