Question

In a right angle triangle abc right angle is at


c. bc+ca=23cm and bc-ca=7cm ,then find sin a and tanb

Answer 1

Hey there!

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In a right angle triangle ABC, right angle is at C.

BC + C̶A̶ + BC - C̶A̶ = 23 + 7 =30

2BC = 30

BC = $\frac{30}{2} = 15$

BC + CA = 23

15 + CA = 23

CA = 23 - 15 = 8

According to Pythagoras Theorem,

${ab}^{2} = {ac}^{2} + {bc}^{2} \\ {ab}^{2} = {8}^{2} + {15}^{2} \\ {ab}^{2} = 64 + 225 = 289 = {17}^{2} \\ {ab} = 17cm \\ \\ hence \\ \: \\ \sin( a ) = \frac{bc}{ab} = \frac{15}{17} \\ \tan( b) = \frac{ac}{bc} = \frac{8}{15}$