Question

Given: ∆SPM, PK⊥ SM SP = 25, SM = 28, PK = 9 Find: m∠S, m∠M, PM

Answer 1

Answer:

∠S = 21.10°

∠M = 79.45°

PM = 9.16

Step-by-step explanation:

Here Pythagorean theorem and trigonometry suffice to solve our problem.

From the Pythagorean theorem we get:

$KM^2+9^2=PM^2$ and

$(25-KM)^2+9^2=25^2.$

We solve for $KM$ in the second equation and get:

$(25-KM)=\sqrt{25^2-9^2}$

$\therefore KM=25-\sqrt{25^2-9^2} =\boxed{1.68}$

Now since

$SK+KM=25\\\\ SK=\boxed{23.32}$

Therefore

${\angle}S=Tan^{-1}(\frac{PK}{SK}) = Tan^{-1}(\frac{9}{23.32})=21.10^o$

and

${\angle}M=Tan^{-1}(\frac{PK}{KM}) = Tan^{-1}(\frac{9}{1.68})=79.45^o.$

And finally again from the Pythagorean theorem:

$PM^2=PK^2+KM^2=9^2+1.68^2$

$\therefore PM=\sqrt{9^2+1.68^2} =9.16.$

Thus,

∠S = 21.10°

∠M = 79.45°

PM = 9.16.