मराठी

In the following figure, ΔABC is right-angled at B, ΔBSC is right-angled at S and ΔBRS is right-angled at R, AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°. Find (i) tan x° - Mathematics

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प्रश्न

In the following figure, ΔABC is right-angled at B, ΔBSC is right-angled at S and ΔBRS is right-angled at R, AB = 18 cm, BC = 7.5 cm, RS = 5 cm, ∠BSR = x° and ∠SAB = y°. Find (i) tan x°, (ii) sin y°, (iii) cos y°.

बेरीज
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उत्तर

Given: From the figure: ΔABC is right-angled at B, ΔBRS is right-angled at R; AB = 18 cm, BC = 7.5 cm, RS = 5 cm, BR = 6 cm, ∠BSR = x°, ∠SAB = y°.

Step-wise calculation:

1. For x = ∠BSR (triangle BRS, right-angled at R):

In ΔBRS, opposite to angle at S is BR = 6 and adjacent is RS = 5.

tan x = `"Opposite"/"Adjacent"`

= `"BR"/"RS"`

= `6/5`

2. For y = ∠SAB (note R lies on AB, so ∠SAB = ∠SAR; triangle ARS is right-angled at R):

AR = AB – BR

= 18 – 6

= 12

RS = 5, so `AS = sqrt(AR^2 + RS^2)`

= `sqrt(12^2 + 5^2)`

= `sqrt(144 + 25)`

= `sqrt(169)`

= 13

sin y = `"Opposite"/"Hypotenuse"`

= `"RS"/"AS"`

= `5/13`

cos y = `"Adjacent"/"Hypotenuse"`

= `"AR"/"AS"`

= `12/13`

(i) tan x = `6/5`. 

(ii) sin y = `5/13`. 

(iii) cos y = `12/13`.

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पाठ 17: Trigonometric Ratios - Exercise 17A [पृष्ठ ३६०]

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नूतन Mathematics [English] Class 9 ICSE
पाठ 17 Trigonometric Ratios
Exercise 17A | Q 15. | पृष्ठ ३६०
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