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प्रश्न
The rod AC of TV disc antenna is fixed at right angles to wall AB and a rod CD is supporting the disc as shown in the figure. If AC = 1.5 m long and CD = 3 m, find (i) tan θ (ii) sec θ + cosec θ.
योग
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उत्तर
AD = `sqrt(9 - 2.25)`
= `sqrt(6.75)`
= `(3sqrt(3))/2`
tan θ = `"CA"/"AD"`
= `1.5/(3sqrt(3)) xx 2/1`
= `1/sqrt(3)`
sec θ + cosec θ = `"CD"/"AD" + "CD"/"CA"`
= `3[(1 xx 2)/(3sqrt(3)) + 1/1.5]`
= `3[2/(3sqrt(3)) + 2/3]`
= `6[(1 + sqrt(3))/(3sqrt(3))]`
= `(2(sqrt(3) + 1))/sqrt(3)`
= `2/3 (3 + sqrt(3))`
(i) tan θ = `1/sqrt(3)`
(ii) sec θ + cosec θ = `2/3 (3 + sqrt(3))`
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