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In ΔPQR, ∠P and ∠Q are complimentary angles. If sinQ = 4/5 , cosP + cosQ =

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complimentary means

P + Q = 90

P = 90 - Q

Cos(p) = cos(90-Q) = sinQ = 4/5 ( given)

You can find CosQ from the triangle of Q

CosQ = sqrt(5^2 - 4^2)/5 = 3/5


Required = CosP + CosQ = 4/5 + 3/5 = 7/5
User Sabareesh Kkanan
by
8.0k points
4 votes

Answer:

CosP +CosQ =
(7)/(5) .

Explanation:

Given : In ΔPQR, ∠P and ∠Q are complimentary angles. If sinQ = 4/5 .

To find : cosP + cosQ =

Solution : We have given that ∠P and ∠Q are complimentary angles.

P + Q = 90 .

Q = 90 - P.

Then sin( 90 -p ) =
(4)/(5) .

As we know that sin( 90 -p ) = CosP =
(4)/(5) .

We have SinQ =
(4)/(5) .

CosQ =
\sqrt{1 - Sin^(2)Q }

Plugging the value of SinQ

CosQ =
\sqrt{1 - ((4)/(5)) ^(2) }.

CosQ =
\sqrt{1 - ((16)/(25))}.

CosQ =
\sqrt{(9)/(25) }.

CosQ =
(3)/(5) .

CosP +CosQ =
(4)/(5) +
(3)/(5).

CosP +CosQ =
(7)/(5) .

Therefore, CosP +CosQ =
(7)/(5) .

User Sovon
by
8.2k points

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