In $△$ PQR right angled at Q,PR+QR=25 cm and PQ=5 cm. Find all the sides

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Solution

$P R + Q R = 25$ (given) and $P R = 25 - Q R$

$5^{2} + Q R^{2} = P R^{2}$ (Pythagorean theorem)

Then substitute PR to get this equation = $5^{2} + Q R^{2} = (25 - Q R)^{2}$

Solve for QR

$5^{2} + Q R^{2} = 625 - 50 Q R + Q R^{2}$ ( $Q R^{2}$ cancels out)

$5^{2} = 625 - 50 Q R$

$- 600 = - 50 Q R$

$Q R = 12$

Solve for PR using original equation

$P R = 13$

Now draw the triangle on your paper for help solving the next step
$sin = \frac{o p p}{h y p} \Rightarrow sin (P) = \frac{12}{13}$
$cos = \frac{a d j}{h y p} \Rightarrow cos (P) = \frac{5}{13}$
$tan = \frac{o p p}{a d j} \Rightarrow tan (P) = \frac{12}{5}$

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