Question

In angle OPQ, right angled at P, OP = 7cm and OQ -PQ = 1 cm. In the adjoining figure, determine the values of sin Q and cos Q

Answer 1

Step-by-step explanation:

in triangle o p q o cube minus b cube is equal to 1 cm cube is equal to 1 + PQ by Pythagoras Theorem Oka square is equal to PQ square + c square one plus b cube ka whole square is equal to PQ square + 7 square 1 + PQ square + 2 PQ is equal to PQ square + 41 + 2 square + 2 PQ - PQ square - 49 is equal to 0 1 + 2 PQ - 14 is equal to 2 PQ - 48 is equal to zero to PQ is equal to 48 PQ is equal to 24 is equal to 1 + PQ is equal to 1 + 24 kg equal to 25 now sin theta is equal to 1 upon sin theta is equal to 7 upon 25 cos theta is equal to P Q upon you is equal to 24 upon 25

Answer 2

Step-by-step explanation:

let OQ = x & PQ = y

OQ - PQ =1 (GIVEN)

í.e. x - y = 1 ...... (1)

In triangle OPQ, angle OPEN = 90'

OQ^2 = OP^2 + PQ^2 ....... (P.G.T)

OQ^2 - PQ^2 = 7^2

(OQ - PQ)(OQ+PQ) = 49

OQ+PQ=49

x+y=49 ........ (2)

add (1) & (2)

we get x = 25

put x = 25 in (1)

we get y = 24

Therefore, sin Q = OP/OQ = 7/25

cos Q = PQ/OQ = 24/25