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Look at the figures given below :

Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.
Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.
Look at the figures given below. Trigonometrical Ratios, R.S. Aggarwal Mathematics Solutions ICSE Class 9.

From these figures, write down the values of :

(i) sin x

(ii) tan x

(iii) sec x

(iv) cos y

(v) cot y

(vi) cosec y

(vii) sin z

(viii) cos z

(ix) tan z

Trigonometrical Ratios

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Answer

(i) sin x = perpendicularhypotenuse=qr\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{q}{r}

(ii) tan x = perpendicularbase=qp\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{q}{p}

(iii) sec x = hypotenusebase=rp\dfrac{\text{hypotenuse}}{\text{base}} = \dfrac{r}{p}

(iv) cos y = basehypotenuse=bn\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{b}{n}

(v) cot y = baseperpendicular=bm\dfrac{\text{base}}{\text{perpendicular}} = \dfrac{b}{m}

(vi) cosec y = hypotenuseperpendicular=nm\dfrac{\text{hypotenuse}}{\text{perpendicular}} = \dfrac{n}{m}

(vii) sin z = perpendicularhypotenuse=un\dfrac{\text{perpendicular}}{\text{hypotenuse}} = \dfrac{u}{n}

(viii) cos z = basehypotenuse=kn\dfrac{\text{base}}{\text{hypotenuse}} = \dfrac{k}{n}

(ix) tan z = perpendicularbase=uk\dfrac{\text{perpendicular}}{\text{base}} = \dfrac{u}{k}

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