Question

Q9)An archaeological site, which was
used for quarrying limestone was:
O (a) Chesowanja, Kenya
O (b) Schoningen, Germany
O (c) Dolni Vestonice, Czech Republic
O (d) Neander Valley, Germany​

Answer 1

The trigonometric ratios are defined as follows:-

1. Sine.

$sin ∅ = \frac{opposite}{hypotenuse}$

2. cosine.

$cos ∅= \frac{adjacent}{hypotenuse}$

3. Tangent.

$tan ∅ = \frac{opposite}{adjacent}$

4. cosecant

$csc ∅=\frac{hypotenuse}{opposite}$

5. secant

$sec ∅ = \frac{hypotenuse}{adjacent}$

6. cotangent

$cot ∅=\frac{adjacent}{opposite}$

Explanation:

Given,

To find all the six trigonometric ratio of ∅ use above mentioned formula for trigonometric ratio before it find the value of adjacent.

Step 1 of 2

Given, hypotenuse=10 and opposite =3

therefore, By Pythagoras theorem

(opposite)²+(adjacent)²=(hypotenuse)²

(3)² +(adjacent)²=(10)²

(adjacent)²= (10)²-(3)²

(adjacent)²=100-9

(adjacent)²=91

$adjacent= \sqrt{91}$

step 2 of 2

1. $sin ∅ = \frac{opposite}{hypotenuse}$

$sin ∅ = \frac{3}{10}$

2 $.cos ∅= \frac{adjacent}{hypotenuse}$

$cos ∅=\frac{ \sqrt{91} }{10}$

3 $.tan ∅ = \frac{opposite}{adjacent}$

$tan ∅ =\frac{3}{\sqrt{91}}$

4 $.csc ∅=\frac{hypotenuse}{opposite}$

$csc ∅=\frac{10}{3}$

5. $sec ∅ = \frac{hypotenuse}{adjacent}$

$sec ∅ = \frac{10}{\sqrt{91}}$

6. $cot ∅=\frac{adjacent}{opposite}$

$cot ∅=\frac{\sqrt{91}}{3}$

Final Answer:

The value of all six trigonometric ratios are:-

1. $sin ∅ = \frac{3}{10}$
2. $cos ∅=\frac{ \sqrt{91} }{10}$
3. $tan ∅ =\frac{3}{\sqrt{91}}$
4. $csc ∅=\frac{10}{3}$
5. $sec ∅ = \frac{10}{\sqrt{91}}$
6. $cot ∅=\frac{\sqrt{91}}{3}$