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Exercise 3.3 · Q2

Q.(57, 267)\left(\dfrac57,\ \dfrac{2\sqrt6}{7}\right) is a point on the terminal side of an angle θ\theta in standard position. Determine the trigonometric function values of angle θ\theta.

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With P(x,y)=(57,267)P(x,y)=\left(\dfrac57,\dfrac{2\sqrt6}{7}\right) on the terminal side of θ\theta, first find r=OPr=OP, then apply the coordinate definitions of §3.4.1 directly.

Step 1. Identify x,yx,y. x=57x=\dfrac57, y=267y=\dfrac{2\sqrt6}{7}.

Step 2. Compute r=x2+y2r=\sqrt{x^2+y^2}.

x2=2549,y2=4⋅649=2449,x2+y2=25+2449=4949=1.x^2=\frac{25}{49},\qquad y^2=\frac{4\cdot6}{49}=\frac{24}{49},\qquad x^2+y^2=\frac{25+24}{49}=\frac{49}{49}=1.

So r=1=1r=\sqrt1=1 — the given point already lies on the unit circle.

Step 3. Sine and cosine. sin⁡θ=yr=267,cos⁡θ=xr=57.\sin\theta=\dfrac{y}{r}=\dfrac{2\sqrt6}{7},\qquad \cos\theta=\dfrac{x}{r}=\dfrac57.

Step 4. Tangent and cotangent.

tan⁡θ=yx=26/75/7=265,cot⁡θ=xy=5/726/7=526=5612  (rationalising, multiplying by 6/6).\tan\theta=\frac{y}{x}=\frac{2\sqrt6/7}{5/7}=\frac{2\sqrt6}{5},\qquad \cot\theta=\frac{x}{y}=\frac{5/7}{2\sqrt6/7}=\frac{5}{2\sqrt6}=\frac{5\sqrt6}{12}\ \ (\text{rationalising, multiplying by }\sqrt6/\sqrt6).

Step 5. Cosecant and secant.

cosec⁡θ=ry=126/7=726=7612,sec⁡θ=rx=15/7=75.\operatorname{cosec}\theta=\frac{r}{y}=\frac{1}{2\sqrt6/7}=\frac{7}{2\sqrt6}=\frac{7\sqrt6}{12},\qquad \sec\theta=\frac{r}{x}=\frac{1}{5/7}=\frac75.

Step 6. Check. sin⁡2θ+cos⁡2θ=2449+2549=1\sin^2\theta+\cos^2\theta=\dfrac{24}{49}+\dfrac{25}{49}=1 ✓, confirming the six values are mutually consistent.

✓Final answer

sin⁡θ=267, cos⁡θ=57, tan⁡θ=265, cosec⁡θ=7612, sec⁡θ=75, cot⁡θ=5612\sin\theta=\dfrac{2\sqrt6}{7},\ \cos\theta=\dfrac57,\ \tan\theta=\dfrac{2\sqrt6}{5},\ \operatorname{cosec}\theta=\dfrac{7\sqrt6}{12},\ \sec\theta=\dfrac75,\ \cot\theta=\dfrac{5\sqrt6}{12}.

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