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cot θ = 7/8 | Find all other trigonometric ratios | Trigonometry Numerical

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We have already studied the basics of Trigonometry and the formulas of Trigonometry in the previous notes about Trigonometry. If you haven't read the notes, read it now

Question: If  cotθ=78\ cot \theta = {7 \over 8}; then find all other trigonometric ratios.

Answer

Given that, cotθ=78cot \theta = {7 \over 8}

As shown in the figure below, let's assume that we have a right triangle ABC\triangle ABC, where mB=90°m \angle B = 90 \degree

triangle-abc-with-angle-b-as-90-degree

Now from the formula of cotθcot \theta, that we learned in this notes, we know that:

cot θ=Adjacent side to θOpposite side to θcot \ \theta = {\text{Adjacent side to } \theta \over \text{Opposite side to } \theta}

Putting values in this equation and solving further,

cotθ=BCAB78=BCAB\begin{aligned} \therefore cot \theta &= {BC \over AB} \\ \therefore {7 \over 8} &= {BC \over AB} \\ \end{aligned}

From the equation above, let
AB=8xAB = 8 \sdot x
BC=7xBC = 7 \sdot x

Putting the values of ABAB and BCBC in the figure with ABC\triangle ABC

triangle-abc-with-side-values

Now,
By Pythagoras theorem

AC2=AB2+BC2=(8x)2+(7x)2=64x2+49x2AC2=113x2AC=113x2=113x\begin{aligned} AC^2 &= AB^2 + BC^2 \\ &= (8x)^2 + (7x)^2 \\ &= 64x^2 + 49x^2 \\ \therefore AC^2 &= 113x^2 \\ \therefore AC &= \sqrt {113x^2} \\ &= \sqrt {113} \sdot x \end{aligned}

Now, that we have the values of ABAB, BCBC and ACAC we can find the values of sin θsin \ \theta, cos θcos \ \theta, tan θtan \ \theta, cosec θcosec \ \theta and sec θsec \ \theta by applying the formulas that we learnt in the basics of trigonometry

For sin θsin \ \theta,

sin θ=Opposite side to θHypotenuse=ABAC=8x113x=8x113xsin θ=8113\begin{aligned} sin \ \theta &= {\text {Opposite side to } \theta \over Hypotenuse} \\ \\ &= {AB \over AC} \\ \\ &= {8x \over \sqrt {113} \sdot x} \\ \\ &= {8 \cancel{x} \over \sqrt {113} \sdot \cancel{x}} \\ \\ & \boxed {\therefore sin \ \theta = {8 \over \sqrt {113}}} \\ \end{aligned}

For cos θcos \ \theta,

cos θ=Adjacent side to θHypotenuse=BCAC=7x113x=7x113xcos θ=7113\begin{aligned} cos \ \theta &= {\text {Adjacent side to } \theta \over Hypotenuse} \\ \\ &= {BC \over AC} \\ \\ &= {7x \over \sqrt {113} \sdot x} \\ \\ &= {7 \cancel{x} \over \sqrt {113} \sdot \cancel{x}} \\ \\ & \boxed {\therefore cos \ \theta = {7 \over \sqrt {113}}} \\ \end{aligned}

For tan θtan \ \theta,

tan θ=sin θcos θ=1cot θ=178=87tan θ=87\begin{aligned} tan \ \theta &= {sin \ \theta \over cos \ \theta} \\ \\ &= {1 \over cot \ \theta} \\ \\ &= {1 \over {7 \over 8}} \\ \\ &= {8 \over 7} \\ \\ & \boxed {\therefore \tan \ \theta = {8 \over 7}} \end{aligned}

For sec θsec \ \theta,

sec θ=1cos θ=17113=1137sec θ=1137\begin{aligned} sec \ \theta &= {1 \over cos \ \theta} \\ \\ &= {1 \over {7 \over \sqrt {113}}} \\ \\ &= {\sqrt {113} \over 7} \\ \\ &\boxed {\therefore sec \ \theta = {\sqrt {113} \over 7}} \end{aligned}

For cosec θcosec \ \theta,

cosec θ=1sin θ=18113=1138cosec θ=1138\begin{aligned} cosec \ \theta &= {1 \over sin \ \theta} \\ \\ &= {1 \over {8 \over \sqrt {113}}} \\ \\ &= {\sqrt {113} \over 8} \\ \\ &\boxed {\therefore cosec \ \theta = {\sqrt {113} \over 8}} \end{aligned}

If you have doubts, please leave a comment below.

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