Question

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Answer 1

Answer:

We know, that the triangle ABC is right angled at C

i.e. ∠C = 90°.

Then, by angle sum property of a triangle:

∠A + ∠B + ∠C = 180°

∠A + ∠B + 90° = 180°

∠A + ∠B = 180° - 90°

∠A + ∠B = 90°

Then, cos(A+B) = cos(90°) = 0

hence, the value of cos(A+B) is 0.

Answer 2

Answer:

Let displacement in nth second

sn = u + \frac{1}{2} a(2n - 1)sn=u+21a(2n−1)

s1 = 0 + \frac{1}{2} a(2p - 1)..........1sts1=0+21a(2p−1)..........1st

and

s2 = 0 + \frac{1}{2} a(2(p + 1) - 1).......2nds2=0+21a(2(p+1)−1).......2nd

And

s = 0 + \frac{1}{2} ap {}^{2} ..........3rds=0+21ap2..........3rd

Now adding equation 1st and 2nd

\begin{lgathered}s1 + s2 = \frac{1}{2} a(2p - 1 + 2p + 1) \\\end{lgathered}s1+s2=21a(2p−1+2p+1)

= 2ap=2ap

Substituting the value of a from equateqution 3rd

s1 + s2 =2 \times \frac{2s}{ {p}^{2} } \times ps1+s2=2×p22s×p

s = \frac{p}{4} (s1 + s2)s=4p(s1+s2)

I hope it will help you a lot