A ladder 10 m long,leans against a vertical wall at an angle of 70° to the ground.if the ladder slips down the wall 4m,find,correct to 2 significant figure
(a) the new angle which the ladder makes with the ground

(b) the distance the ladder slipped back on the ground from it's original position

For now, focus only on triangle ABC. The ladder is segment AC = 10. We first need to find the length of [tex]AB = h_1[/tex] which is the initial height of the ladder.

sin(angle) = opposite/hypotenuse

sin(70) = h/10

h = 10*sin(70)

h = 9.396926 approximately

Subtract off 4 since the ladder slips 4 meters down the wall

h-4 = 9.396926-4

h-4 = 5.396926

which is the new height the ladder reaches. The hypotenuse stays the same

sin(angle) = opposite/hypotenuse

sin(theta) = 5.396926/10

theta = arcsin(5.396926/10)

theta = 32.662715

theta = 33 degrees when rounding to 2 significant figures

This is the value of [tex]\theta_2[/tex] in the diagram below.

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We'll use the cosine rule with the old theta value [tex]\theta_1[/tex]

cos(angle) = adjacent/hypotenuse

cos(70) = x/10

x = 10*cos(70)

x = 3.420201 is the approximate distance the foot of the ladder is from the wall. This is before the ladder slips.

After the ladder slips, we use the new angle value [tex]\theta_2[/tex]

cos(angle) = adjacent/hypotenuse

cos(32.662715) = x/10

x = 10*cos(32.662715)

x = 8.418622

Subtract the two x values

8.418622-3.420201 = 4.998421

which gives the approximate distance the foot of the ladder moved (the distance from point C to point E in the diagram)

This rounds to 5.0 or simply 5 when rounding to 2 significant figures.