You can do this by drawing one line through parallel to PQS to meet RQ at T
Now calculate length of RT:-
cos 70 = RT / 70 giving RT = 23.94m
sin 70 = ST/70 giving ST = 65.78 m
draw a line from S perpendicular to PQ to meet PQ at U.
PU = 110 - 65.78 = 44.22 m
tan 50 = SU / 44.22 giving SU = 52.70 m
TQ = SU = 52.70 m
So x = TQ + RT = 52.70 + 23.94 = 76.6 m to 1 dec place.
Answer:
In group A, they have at least one 90 degree angle. In Group B, they do not have any 90 degree angles.
12x-3y=21
4x-y=7
I hope this helps
We know that
If the scalar product of two vectors<span> is zero, both vectors are </span><span>orthogonal
</span><span>A. (-2,5)
</span>(-2,5)*(1,5)-------> -2*1+5*5=23-----------> <span>are not orthogonal
</span><span>B. (10,-2)
</span>(10,-2)*(1,5)-------> 10*1-2*5=0-----------> are orthogonal
<span>C. (-1,-5)
</span>(-1,-5)*(1,5)-------> -1*1-5*5=-26-----------> are not orthogonal
<span>D. (-5,1)
</span>(-5,1)*(1,5)-------> -5*1+1*5=0-----------> are orthogonal
the answer is
B. (10,-2) and D. (-5,1) are orthogonal to (1,5)
