15 cot A = 8 | Find the value of sin A and sec A | Trigonometry Numerical

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Question: Given 15 cot A = 8, find sin A and sec A

In problems like these, make sure you know all the formulas mentioned in the notes basics of trigonometry by heart. This will help you gain speed and confidence in solving problems like these.

Solution:

Consider ABC\triangle ABC, where mBm \angle B is 90°90 \degree.

15-cot-A-question-triangle-ABC fig. 1

15cot A=8 cot A=815(1)From fig. 1,cotA=Adjacent side to AOpposite side to AcotA=ABBCReplacing this in equation (1) ABBC=815(2)[from figure]\begin{aligned} &15 \sdot \cot \ A = 8 \\ \therefore \ &\cot \ A = {8 \over 15} \quad --- (1)\\ \\ From \ &fig. \ 1, \\ \cot A &= {\text{Adjacent side to } \angle A \over \text{Opposite side to } \angle A} \\ \cot A &= {AB \over BC} \\ \\ Replacing \ &this \ in \ equation \ (1) \\ \therefore \ &{AB \over BC} = {8 \over 15} \quad --- (2) \quad \text{[from figure]} \end{aligned}

From equation (2), we can say that AB=8xAB = 8x and BC=15xBC = 15x.

Wondering why we have taken "8x" and "15x" here ?

Check this video for understanding it.

Applying Pythagoras theorem on ABC\triangle ABC to find the value of ACAC, we get

AC2=AB2+BC2=(8x)2+(15x)2=64x2+225x2 AC2=289x2 AC2=(17x)2 AC=17x(3)\begin{aligned} AC^2 &= AB^2 + BC^2 \\ &= (8x)^2 + (15x)^2 \\ &= 64x^2 + 225x^2 \\ \therefore \ AC^2 &= 289x^2 \\ \therefore \ AC^2 &= (17x)^2 \\ \therefore \ AC &= 17x \quad ---(3) \end{aligned}

Now,

sin A=Opposite side to AHypotenusesin A=BCAC=15x17x sin A=1517\begin{aligned} \sin \ A &= {\text{Opposite side to } \angle A \over \text {Hypotenuse}} \\ \sin \ A &= {BC \over AC} \\ &= {15x \over 17x} \\ &\boxed {\therefore \ \sin \ A = {15 \over 17}} \end{aligned}

sec A=HypotenuseAdjacent side to A=ACAB=17x8x sec A=178\begin{aligned} \sec \ A &= {Hypotenuse \over \text{Adjacent side to } \angle A} \\ &= {AC \over AB} \\ &= {17x \over 8x} \\ &\boxed {\therefore \ \sec \ A = {17 \over 8}} \end{aligned}
Without doing more calculation on paper, can you find the values of tan A, cos A and cosec A? Let us know your answers in comments below and also how you got the values without calculation.

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