Given data, cos(A - B) = cosAcosB + sinAsinB
<span>let, A=60' and B=30' ( here the ' sign bears degree) </span>
<span>L.H.S = cos(A - B) </span>
<span>=cos (60'-30) ( using value of A and B ) </span>
<span>=cos30' </span>
<span>= sqrt3/2 </span>
<span>R.H.S= cosAcosB + sinAsinB </span>
<span>=cos60' cos30' + sin60' sin30' </span>
<span>= 1/2* sqrt3/2+ sqrt3/2*1/2 </span>
<span>= sqrt3/4 + sqrt3/4 </span>
<span>=2 sqrt3/4 </span>
<span>= sqrt3/2 </span>
<span>so L.H.S =R.H.S or cos(A - B) = cosAcosB + sinAsinB</span>
Answer:
The answer is below
Step-by-step explanation:
The question is not complete, what are the coordinates of point Q and R. But I would show how to solve this.
The location of a point O(x, y) which divides line segment AB in the ratio a:b with point A at (
) and B(
) is given by the formula:

If point Q is at (
) and S at (
) and R(x, y) divides QS in the ratio QR to RS is 3:5, The coordinates of R is:

Let us assume Q(−9,4) and S(7,−4)

I believe the answer would be the second option. The equations given above when graph would intersect only once with each other. They intersect at only point (0,6/5). These are the values of x and y that will agree to the equation. Two equations in a set will always intersect at only one point.
<u><em>Answer:</em></u>
AC = 10sin(40°)
<u><em>Explanation:</em></u>
The diagram representing the question is shown in the attached image
Since the given triangle is a right-angled triangle, we can apply the special trig functions
<u>These functions are as follows:</u>
sin(θ) = opposite / hypotenuse
cos(θ) = adjacent / hypotenuse
tan(θ) = opposite / adjacent
<u>Now, in the given diagram:</u>
θ = 40°
AC is the side opposite to θ
AB = 10 in is the hypotenuse
<u>Based on these givens</u>, we will use the sin(θ) function
<u>Therefore:</u>

The answer is 1 because you are the only one in the garden. The rest of the people are in the yard.