Question

In $\Delta OPQ$, right-angled at P, OQ-PQ = 1 cm and OP = 7 cm. Determine the values of sinQ and cosQ .

Answer 1

Hint: Sine function of an angle is the ratio between the opposite side length to that of the hypotenuse. The cosine function of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.

Complete step-by-step answer:

It is given that

OQ-PQ = 1 cm and OP = 7 cm

Let us assume that PQ = x

OQ-PQ = 1

OQ = 1+PQ

OQ = 1 + x

In right triangle OPQ, Using Pythagoras Theorem

\[{{\left( \text{Hypotenuse} \right)}^{2}}={{\left( \text{Height} \right)}^{2}}+{{\left( \text{Base} \right)}^{2}}\]

${{\left( OQ \right)}^{2}}={{\left( OP \right)}^{2}}+{{\left( PQ \right)}^{2}}$

Now put all the values in the above equation, we get

${{\left( 1+x \right)}^{2}}={{7}^{2}}+{{x}^{2}}$

Expanding the term on the left side by using ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , we get

${{1}^{2}}+{{x}^{2}}+2\times 1\times x=49+{{x}^{2}}$

Rearranging the terms, we get

$1+{{x}^{2}}+2x=49+{{x}^{2}}$

$1+{{x}^{2}}+2x-49-{{x}^{2}}=0$

Cancelling the term ${{x}^{2}}$ on the left side, we get

$2x-48=0$

$2x=48$

Dividing both sides by 2, we get

$x=\dfrac{48}{2}=24$

Hence the required length of side PQ $=x=24$cm

Therefore the length of side OQ $=1+x=1+24=25$cm

The basic definitions of the trigonometric functions are

The sine function of an angle is the ratio between the opposite side length to that of the hypotenuse.

$\sin Q=\dfrac{\text{side opposite to angle Q}}{\text{Hypotenuse}}=\dfrac{OP}{OQ}=\dfrac{7}{25}$

The cosine function of an angle is the ratio of the length of the adjacent side to the length of the hypotenuse.

$\cos Q=\dfrac{\text{side adjacent to angle Q}}{\text{Hypotenuse}}=\dfrac{PQ}{OQ}=\dfrac{24}{25}$

Note: It should be noted that the $\cos Q$ can also be represented in terms of sine that is $\cos Q=\sqrt{1-{{\sin }^{2}}Q}$ .

Hence $\cos Q=\sqrt{1-{{\left( \dfrac{7}{25} \right)}^{2}}}=\sqrt{1-\dfrac{49}{625}}=\sqrt{\dfrac{625-49}{625}}=\sqrt{\dfrac{576}{625}}=\sqrt{{{\left( \dfrac{24}{25} \right)}^{2}}}=\dfrac{24}{25}$