Answer to Question #233070 in Trigonometry for dizzo

Question #233070

(a) Find the value of m if the angle between a(3,m) and b (2,-1)is 360 degrees

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(b) Find the longest side of right triangle whose other sides is:(√2+1)cm and (√2-1)cm long



1
Expert's answer
2021-09-06T13:31:16-0400

Solution:

(a)

Given:  a(3,m) and b (2,-1)

Using dot product formula of vectors:

a.b=abcosθa.b=|a||b|\cos \theta

3(2)+m(1)=32+m2×22+(1)2×cos(360°)6m=(9+m2)5×(1)(6m)2=5(9+m2)36+m212m=45+5m24m2+12m+9=04m2+6m+6m+9=0(4m+3)(2m+3)=0m=34,32\\\Rightarrow 3(2)+m(-1)=\sqrt{3^2+m^2}\times\sqrt{2^2+(-1)^2}\times \cos(360\degree) \\\Rightarrow 6-m=\sqrt{(9+m^2)5}\times(1) \\\Rightarrow (6-m)^2=5(9+m^2) \\\Rightarrow 36+m^2-12m=45+5m^2 \\\Rightarrow 4m^2+12m+9=0 \\\Rightarrow 4m^2+6m+6m+9=0 \\\Rightarrow (4m+3)(2m+3)=0 \\\Rightarrow m=-\dfrac34,-\dfrac32

(b)

Two sides are :p=(√2+1)cm and b=(√2-1) cm

We need to find the Hypotenuse (h).

Using Pythagoras theorem,

h2=p2+b2h^2=p^2+b^2

h2=(2+1)2+(21)2h2=2+1+22+2+122h2=6h=6 cm\Rightarrow h^2=(\sqrt2+1)^2+(\sqrt2-1)^2 \\\Rightarrow h^2=2+1+2\sqrt2+2+1-2\sqrt2 \\\Rightarrow h^2=6 \\\Rightarrow h=\sqrt6\ cm


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