Introduction to Trigonometry | Exercise 8.1

Question 4

Given 15cot⁡A=815 \cot A = 8, find sin⁡A\sin A and sec⁡A\sec A.

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Solution
Understand the Question
  • Given 15cot⁡A=8  ⟹  cot⁡A=815=AdjacentOpposite15 \cot A = 8 \implies \cot A = \dfrac{8}{15} = \dfrac{\text{Adjacent}}{\text{Opposite}}.
  • In a right-angled triangle with reference to angle AA, let the adjacent side be 8k8k and the opposite side be 15k15k, where kk is a positive number.
  • Use the Pythagoras theorem to find the hypotenuse, then calculate the required trigonometric ratios:
    • sin⁡A=OppositeHypotenuse\sin A = \dfrac{\text{Opposite}}{\text{Hypotenuse}}
    • sec⁡A=HypotenuseAdjacent\sec A = \dfrac{\text{Hypotenuse}}{\text{Adjacent}}

Step 1 · Find the Hypotenuse

Consider right-angled triangle ABC\text{ABC} right-angled at B\text{B}.Diagram 1

Given

15cot⁡A=8cot⁡A=815\begin{aligned} 15 \cot A &= 8 \\[0.6em] \cot A &= \dfrac{8}{15} \end{aligned}

cot⁡A=Adjacent sideOpposite side=ABBC=815\cot A = \dfrac{\text{Adjacent side}}{\text{Opposite side}} = \dfrac{\text{AB}}{\text{BC}} = \dfrac{8}{15}

Let AB=8k\text{AB} = 8k and BC=15k\text{BC} = 15k, where kk is a positive number.

By Pythagoras theorem in ΔABC\Delta \text{ABC}

AC2=AB2+BC2=(8k)2+(15k)2=64k2+225k2=289k2AC=289k2=17k\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ &= (8k)^2 + (15k)^2 \\ &= 64k^2 + 225k^2 \\ &= 289k^2 \\ \text{AC} &= \sqrt{289k^2} = 17k \end{aligned}

Step 2 · Find sin⁡A\sin A and sec⁡A\sec A

Now compute the required trigonometric ratios

sin⁡A=Opposite sideHypotenuse=BCAC=15k17k=1517\begin{aligned} \sin A &= \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} \\[0.6em] &= \dfrac{15k}{17k} = \dfrac{15}{17} \end{aligned} sec⁡A=HypotenuseAdjacent side=ACAB=17k8k=178\begin{aligned} \sec A &= \dfrac{\text{Hypotenuse}}{\text{Adjacent side}} = \dfrac{\text{AC}}{\text{AB}} \\[0.6em] &= \dfrac{17k}{8k} = \dfrac{17}{8} \end{aligned}
Answer

sin⁡A=1517,sec⁡A=178\sin A = \dfrac{15}{17}, \quad \sec A = \dfrac{17}{8}

Common Mistakes
  • Ratio Definition Error: Confusing cot⁡A=AdjacentOpposite\cot A = \dfrac{\text{Adjacent}}{\text{Opposite}} with tan⁡A=OppositeAdjacent\tan A = \dfrac{\text{Opposite}}{\text{Adjacent}}, which swaps the values of the opposite and adjacent sides.
  • Reciprocal Mistake for sec⁡A\sec A: Remembering that sec⁡A=1cos⁡A=HypotenuseAdjacent\sec A = \dfrac{1}{\cos A} = \dfrac{\text{Hypotenuse}}{\text{Adjacent}}, not HypotenuseOpposite\dfrac{\text{Hypotenuse}}{\text{Opposite}} (which is cosec⁡A\operatorname{cosec} A).

More questions in Exercise 8.1

Q1

In ΔABC\Delta \text{ABC}, right-angled at B, AB=24 cm\text{AB} = 24\text{ cm}, BC=7 cm\text{BC} = 7\text{ cm}. Determine :

(i) sin⁡A\sin A, cos⁡A\cos A

(ii) sin⁡C\sin C, cos⁡C\cos C

Q2

In Fig. 8.13, find tan⁡P−cot⁡R\tan P - \cot R.

Q3

If sin⁡A=34\sin A = \dfrac{3}{4}, calculate cos⁡A\cos A and tan⁡A\tan A.

Q4

Given 15cot⁡A=815 \cot A = 8, find sin⁡A\sin A and sec⁡A\sec A.

Q5

Given sec⁡θ=1312\sec \theta = \dfrac{13}{12}, calculate all other trigonometric ratios.

Q6

If ∠A\angle A and ∠B\angle B are acute angles such that cos⁡A=cos⁡B\cos A = \cos B, then show that ∠A=∠B\angle A = \angle B.

Q7

If cot⁡θ=78\cot \theta = \dfrac{7}{8}, evaluate :

(i) (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}

(ii) cot⁡2θ\cot^2 \theta

Q8

If 3cot⁡A=43 \cot A = 4, check whether 1−tan⁡2A1+tan⁡2A=cos⁡2A−sin⁡2A\dfrac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A or not.

Q9

In triangle ABC, right-angled at B, if tan⁡A=13\tan A = \dfrac{1}{\sqrt{3}}, find the value of:

(i) sin⁡Acos⁡C+cos⁡Asin⁡C\sin A \cos C + \cos A \sin C

(ii) cos⁡Acos⁡C−sin⁡Asin⁡C\cos A \cos C - \sin A \sin C

Q10

In ΔPQR\Delta \text{PQR}, right-angled at Q, PR+QR=25 cm\text{PR} + \text{QR} = 25\text{ cm} and PQ=5 cm\text{PQ} = 5\text{ cm}. Determine the values of sin⁡P\sin P, cos⁡P\cos P and tan⁡P\tan P.

Q11

State whether the following are true or false. Justify your answer.

(i) The value of tan⁡A\tan A is always less than 1.

(ii) sec⁡A=125\sec A = \dfrac{12}{5} for some value of angle A.

(iii) cos⁡A\cos A is the abbreviation used for the cosecant of angle A.

(iv) cot⁡A\cot A is the product of cot⁡\cot and AA.

(v) sin⁡θ=43\sin \theta = \dfrac{4}{3} for some angle θ\theta.

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