Question

Find out the area of a triangle PQR, if p=6, q=3
and cos(P-Q) = 4/5.
A.
12 sq. units
10 sq. units
C.
9 sq. units
D.
8 sq. units
18​

Answer 1

Given:-P=6,Q=3 and cos (p-q)=4/5

Formula:-A=a×b×sin @/2

we have the value of cos theta,using that we'll find the value of sin theta .

=cos theta =b/h(base/hypotenuse)

sin theta =p/h(perpendicular/hypotenuse)

Also assume a triangle with base =4 and hypotenuse=5 ,

for such a triangle,the perpendicular comes out to be =3

Thus,implying p=3

Implies that sin theta=4/5

Now, putting all the values in the formula,we get

A=a×b×sin theta/2

=3×6×4/5×2

=7.2

Area of a triangle PQR, if p=6, q=3

and cos(P-Q) = 4/5,is equal to 7.2 unit square

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Answer 2

Answer:

The correct answer is -  D. 8 sq. units

Explanation:

From the above question,

They have given :

The area of a triangle PQR, if p=6, q=3 and cos(P-Q) = 4/5.

Use the formula for area of a triangle A = ½*base*height

The area of a triangle PQR can be calculated using the formula

A = ½ * base * height.

In order to calculate the base and the height, the law of cosines can be used.

The law states that base = $(p2 + q2 - 2*p*q*cos(P-Q))^{\frac{1}{2} }$ and

                                height =$(p2 + q2 - 2*p*q*cos(P-Q))^{\frac{1}{2} }$.

Given the values of p=6, q=3 and cos(P-Q) = 4/5, the base and height can be calculated and then used to find the area of the triangle. The area of the triangle PQR is then A = ½*base*height

To calculate the base of the triangle using the law of cosines,

                base = ($p2 + q2 - 2*p*q*cos$(P-Q)$)^{\frac{1}{2} }$

To calculate the height of the triangle using the law of cosines,

                height = ($p2 + q2 - 2*p*q*cos$(P-Q)$)^{\frac{1}{2} }$

To calculate the area of the triangle,

                A = ½*base*height

                    = ½*($6^2 + 3^2 - 2*6*3*(4/5$)$)^{\frac{1}{2} }$

                    = 8 sq. units

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