Question

Given: In ∆ ,∠=°.

If 15 cot A=8, Find sin A

and sec A.​

Answer 1

Answer:

$answer \: is \: in \: attachment$

we fix an acute angle θ, then all right-angled triangles that have θ as one of their angles are similar. So, in all such triangles, corresponding pairs of sides are in the same ratio.

Right angle triangle, sides labelled 'hypotenuse', 'opposite' and 'adjacent.

The side opposite the right angle is called the hypotenuse. We label the side opposite θ as the opposite and the remaining side as the adjacent. Using these names we can list the following standard ratios:

sinθ=oppositehypotenuse,cosθ=adjacenthypotenuse,tanθ=oppositeadjacent.

Allied to these are the three reciprocal ratios, cosecant, secant and cotangent:

cosecθ=hypotenuseopposite,secθ=hypotenuseadjacent,cotθ=adjacentopposite.

These are called the reciprocal ratios as

cosecθ=1sinθ,secθ=1cosθ,cotθ=1tanθ.

The trigonometric ratios can be used to find lengths and angles in right-angled triangles.

Example

Find the value of x in the following triangle.

Right angle triangle, adjacent labelled as x, opposite marked as 3, angle created by the adjacent and hypotenuse marked as 72 degrees.

Solution

We have

x3=cot72∘=1tan72∘.

Hence

x=3tan72∘≈0.97,correct to two decimal places.

Special angles

The trigonometric ratios of the angles 30∘, 45∘ and 60∘ can be easily expressed as fractions or surds, and students should commit these to memory.

Trigonometric ratios of special angles

θ sinθ cosθ tanθ

30∘ 12 3–√2 13–√

45∘ 12–√ 12–√ 1

60∘ 3–√2 12 3–√

Right angle triangle on the left, adjacent marked as 1, hypotenuse marked as root 2, on the right a Right angle triangle, adjacent marked as 1, hypotenuse marked as 2.

Triangles for the trigonometric ratios of special angles.

Answer 2

Given that,

$\rm { \qquad • 15 cot \: A = 8 }$

_________

We have to find,

$\rm { \qquad • sin \: A \: and \: sec \: A }$

_________

Solution,

If $\rm {15 cot \: A = 8 }$, then $\rm { cot \: A = \dfrac{8}{15} }$

We know that ,

$\rm { \qquad • cot \: A = \dfrac{Base}{Perpendicular}}$

Hence,

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

____________

Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2}+{P}^{2} }$

$\rm\leadsto { {H}^{2} = {8}^{2}+{15}^{2} }$

$\rm\leadsto { {H}^{2} = 64+ 225}$

$\rm\leadsto { {H }^{2} = 289}$

$\leadsto\rm\blue { H = \sqrt{289} = 17}$

_____________

$\rm { \qquad • Base = 8 }$

$\rm { \qquad • Perpendicular = 15 }$

$\rm { \qquad • Hypotenuse = 17 }$

$\qquad$

$\rm\red { \leadsto sin \: A = \dfrac{ Perpendicular}{Hypotenuse} =\dfrac{15}{17} }$

$\qquad$

$\rm\red { \leadsto sec \: A = \dfrac{ Hypotenuse}{Base} =\dfrac{17}{8} }$