if in a right angled triangle ABC, LB is a right angle; then find the value of Sin A, cos C find the value of sin A, cos C and tan A. (i) AC = 5 AB = 3 BC=4
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Answer:
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Answer:
question :-
Find the value of the expressions for the given value of x.
(a) -3x + 2 for x = -1
(b) 3x^2 + 2x - 5 for x = - 4
(c) 9x - 5 for x = 3
(d) - 2x^2 + 3x - 2 for x = 2
Solution 1 :-
put the value of x = -1 in the equation
\mathsf{(a) \: - 3 x\: + \: 2}(a)−3x+2
\mathsf{: \longrightarrow \: - 3 \: \times \: - 1 + \: 2}:⟶−3×−1+2
\mathsf{: \longrightarrow \: 3 \: + \: 2}:⟶3+2
\mathsf{: \longrightarrow \: 5}:⟶5
Solution 2 :-
Put the value of x = - 4 in the equation .
\mathsf{(b) \: \:3x^2 + 2x - 5 }(b)3x
2
+2x−5
\mathsf{: \longrightarrow \:3 \: ( - 4)^2 \: + \: 2 \: \times \: - 4 \: - 5 }:⟶3(−4)
2
+2×−4−5
\mathsf{: \longrightarrow \:3 \: \times \: 16 \: - \: 8 \: - \: 5 }:⟶3×16−8−5
\mathsf{: \longrightarrow \:48 \: - \: 8 \: - \: 5 }:⟶48−8−5
\mathsf{: \longrightarrow \:40 \: - \: 5 }:⟶40−5
\mathsf{: \longrightarrow \: 35 }:⟶35
Solution 3 :-
Put the value of x = 3
\mathsf{(c) \: \:9x \: - \: 5 }(c)9x−5
\mathsf{: \longrightarrow \:9 \: \times \: 3\: - \: 5 }:⟶9×3−5
\mathsf{: \longrightarrow \:23 \: - \: 5 }:⟶23−5
\mathsf{: \longrightarrow \: 18}:⟶18
Solution 4 :-
Put the value of x = 2 in the equation .
\mathsf{(d)\:- 2x^2 \: + \: 3x \: - \: 2 }(d)−2x
2
+3x−2
\mathsf{: \longrightarrow \:- 2(2)^2 \: + \: 3 \: \times \: 2 \: - \: 2 }:⟶−2(2)
2
+3×2−2
\mathsf{: \longrightarrow \:- 2 \: \times \: 4\: + \: 3 \: \times \: 2 \: - \: 2 }:⟶−2×4+3×2−2
\mathsf{: \longrightarrow \: - 8\: + \:6 \: - \: 2 }:⟶−8+6−2
\mathsf{: \longrightarrow \:-10 \: + \: 6}:⟶−10+6
\mathsf{: \longrightarrow \:- \: 4}:⟶−4