Question

In the figure 8.15,.∠PQR=90°, ∠PQS=90°, ∠PRQ= and ∠QPS=θ Write the following trigonometric ratios.
(i) sin , cos , tan
(ii) sin θ, cos θ, tan θ

Answer 1

opposite side
opposite angle

Answer 2

Answer:

(i):

$\rm \sin\alpha = \dfrac{PQ}{PR}.\\\cos\alpha = \dfrac{RQ}{PR}.\\\tan\alpha = \dfrac{PQ}{RQ}.$

(ii):

$\rm \sin\theta=\dfrac{QS}{PS}.\\\cos\theta = \dfrac{PQ}{PS}\\\tan\theta = \dfrac{QS}{PQ}.$

Step-by-step explanation:

For a right-angled triangle, the trigonometric ratios are defined as

The sine (sin) of an angle = the ratio of the perpendicular to the hypotenuse of the triangle.

The cosine (cos) of an angle = the ratio of the base to the hypotenuse of the triangle.

The tangent (tan) of an angle = the ratio of perpendicular to the base of the triangle.

For an angle in a right-angled triangle, the side of the triangle which is opposite to that angle is called perpendicular, the side which is adjacent to that angle is called base and the largest side of the triangle is called  hypotenuse.

(i):

In triangle PQR, for the angle $\alpha$,

Perpendicular = PQ.

Base = RQ.

Hypotenuse = PR.

$\rm \sin\alpha = \dfrac{PQ}{PR}.\\\cos\alpha = \dfrac{RQ}{PR}.\\\tan\alpha = \dfrac{PQ}{RQ}.$

(ii):

In triangle PQS, for the angle $\theta$,

Perpendicular = QS.

Base = PQ.

Hypotenuse = PS.

$\rm \sin\theta=\dfrac{QS}{PS}.\\\cos\theta = \dfrac{PQ}{PS}\\\tan\theta = \dfrac{QS}{PQ}.$