Question

6. In the figure, given below, ABC is an isosceles
triangle 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-----​

Answer 1

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.

Answer 2

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