Engineering Mechanics - General Principles - Discussion

From the given diagram, let AB=BC=x and AC=d
x^2 + x^2 = d^2
2(x^2) =d^2
on Substituting the value of d=10
2(x^2) = 100
x^2 = 50
= 2x25
x = 5 x [(2)^(1/2)]
= 5 x 1.141
= 7.07

SOH-CAH-TOA.

SIN(THETA) = OPPOSITE/HYPOTENUSE.
COS(THETA) = ADJACENT/HYPOTENUSE.

TAN(THETA) = OPPOSITE/ADJACENT.

THEREFORE AB = CAH.
AB = COS (THETA) = adj/hyp.
AB = COS 45 = ADJ/10.
AB = 10. COS45.

And, COS45 = 0.707,
AB = 10 . 0.707
AB = 7.07.

AB = Perpendicular; AC = hypotenuse.

Now we know sin 45 = AB/AC.

->> 1/sqrt2 = AB/10.

->> 0.707 = AB/10 (1/sqrt2 = 0.707).

->> AB = 10*0.707.

->> AB = 7.07.

Perpendicular AB
sin(45)= ---------------- = ------
hypotenuse 10

So,

AB = sin(45)x 10 = 0.707x10 = 7.07 Ans.

Sin θ =opposite side/hypotenuse sin(theera) = AB/ AC.
AB=Ac sin45. = 10 sin45= 10 * 0.707.

So, AB = 7.07.

In Given Triangle by using trignomtric relations
sin (theta)= AB/AC=AB/10
we know that sin45=0.707
0.707=AB/10
=>AB=0.707*10

H/B=SIN 45*.

=>AB/AC= SIN 45*.

=>AB/10=0.707106781.

=>AB=0.707106781*10.

=>AB=7.07.

Its very to find out,
Sin 45 = opp/ hy
opp = sin 45 x hy
= 1/sq.root of 2 x 10
opp = 7.07

Without calculator how will you find out sin 45 is 0.707. Did you use any tricks?

Let base and height= x (becoz angle is 45)

So x2+X2= 2X2=100

HENCE X=7.07