मराठी

Find tan θ + cot α, sin θ, cos α. - Mathematics

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प्रश्न

Find tan θ + cot α, sin θ, cos α.

बेरीज
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उत्तर

Given:

AB = 20cm

BC = 15 cm

CD = 6 cm

Step 1: In triangle △ABC

Apply Pythagoras Theorem:

AC2 = AB2 + BC2

AC2 = 202 + 152

AC2 = 400 + 225

AC2 = 625

AC = `sqrt 625` = 25 cm

Step 2: Calculate the trigonometric values for angle θ in △ABC:

`sin θ = "Opposite"/"hypotenuse" = (BC)/(AC) = 15/25 = 3/5`

`cos θ = "adjacent"/"hypotenuse" = (AB)/(AC) = 20/25 = 4/5`

`tan θ = "Opposite"/"adjacent" = (BC)/(AB) = 15/20 = 3/5`

Step 3: Calculate the trigonometric values for angle α in △ACD:

In △ACD, the opposite side to angle α is AB = 20

The adjacent side to angle α is BD = BC + CD = 15 + 6 = 21 

The hypotenuse AD needs to be calculated using the Pythagorean theorem in △ABD   ...(assuming it's right-angled at B)

AD2 = AB2 + BD2 = 202 + 212 = 400 + 441 = 841

AD = `sqrt841` = 29

`cot α = "adjacent"/"opposite" = (BD)/(AB) = 21/20`

`cos α = "adjacent"/"hypotenuse" = (BD)/(AD) = 21/29`

Step 4: Find tan⁡ θ + cot⁡ α:

tan θ + cot α = `3/4 + 21/20`

Find a common denominator (20)

`3/4 + 15/20`

`15/20 + 21/20 = 36/20`

`36/20 = 9/5`

Final Answer:

tan θ + cot α = `9/5`

sin θ = `3/5`

cos α = `21/29`

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पाठ 19: Trigonometry - EXERCISE 19A [पृष्ठ २३१]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
पाठ 19 Trigonometry
EXERCISE 19A | Q I. 8. | पृष्ठ २३१
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