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6. In the figure, given below, ABC is an isoscelestriangle with BC = 8 cm and AB = AC = 5 cm.Find :(i) sin B(ii) tan c(iii) sin? B + cos2 B (iv) tan C - cot B-cm----- |
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Answer» Step-by-step explanation: Here AB = BC = 5 cm and we draw an altitude AD as we know in isosceles TRIANGLE altitude also a median for not equal side of isosceles triangle . So BD = CD = 4 cm ( Given BC = 8 cm ) Now from Pythagoras theorem In ∆ ABD we get AB^2 = AD^2 + BD^2 ,Substitute values and get 5^2 = AD^2 + 4^2 AD^2 = 25 - 16 AD^2 = 9 AD = 3 cm i ) We know : Sin =Opposite/Hypotenuse , So Sin B = ADAB⇒Sin B = 35 ( ANS ) II ) We know : tan θ = Opposite/Adjacent , So tan C = ADCD⇒tan C = 34 ( Ans ) iii ) We know : Cos θ = Adjacent/Hypotenuse , So Cos B= BDAB⇒Cos B= 45So, Sin2 B + Cos^2 B ⇒(35)^2 + (45)^2⇒925 + 1625⇒9 + 1625⇒2525⇒1 ( Ans ) iv ) We know : Cot θ = AdjacentOpposite , So Cot B= BDAD⇒Cot B= 43So, tan C − Cot B ⇒34 − 43⇒9 − 1612⇒−712 ( Ans ) Hope this information will clear your doubts about Trigonometry. |
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