Question

Q1. Draw a trigonometry table.
Q2. In a right angled triangle, if angle A is acute and cot A = 4/3; find the remaining trigonometrical ratios.


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Answer 1

Answer:

Step-by-step explanation:

→ Answer is in attachment.

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Answer 2

Required answers:-

Questions:

Q1. Draw a trigonometry table

Q2. In a right angled triangle, if angle A is acute and cot A= 4/3; find the remaining trigonometrical ratios.

Solution:

Answer1.

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Answer2.

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Given,

★ cot A = 4/3

To find:

★ Remaining trigonometrical ratios

Formulas used:

★ cot A = base/perpendicular → AB/BC

★ sec A = hypotenuse/base → AC/AB

★ tan A = perpendicular/base → BC/AB

★ cosec A = hypotenuse/perpendicular → AC/BC

★ cos A = base/hypotenuse → AB/AC

★ sin A = perpendicular/hypotenuse → BC/AC

Assumptions:

★ Let AB = 4x

★ Let BC = 3x

Step by step explaination:

Using Pythagoras theorem....

As we know

Hypotenuse = Base + Perpendicular

➡ AC² = AB² + BC²

➡ AC² = (4x)² + (3x)²

➡ AC² = 25x²

➡ AC = 5x

Now finding the remaining trigonometrical ratios...

• sin A = perpendicular/hypotenuse

= 3x/5x

= 3/5

• tan A = perpendicular/base

= 3x/4x

= 3/4

• cosec A = hypotenuse/perpendicular

= 5x/3x

= 5/3

• sec A = hypotenuse/base

= 5x/4x

= 5/4

• cos A = base/hypotenuse

= 4x/5x

= 4/5

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Know about the terms:

• The word "Trigonometry" means measurement of triangles.

• For any acute angle in a right-angled triangle; the side opposite to the acute angle is called the perpendicular.

• The side adjacent to it is called the base and the side opposite to the right angled triangle is called hypotenuse.

• The ratio between the lengths of a pair of two sides of a right angled triangle is called a trigonometrical ratio.

Note:

• Each trigonometrical ratio is a real number and has no unit.

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