Question

if 9sinα+40cosα=41 then find all trigonometry ratio​

Answer 1

9 sin θ + 40 cos θ = 41

⇒ 9 sin θ = 41 – 40 cos θ _(i)

Squaring both sides, we get

⇒ 81sin²θ = 1681+1600 cos²θ – 2(41) (40cos θ)

[∵ (a – b)² = a² + b² –2ab]

⇒ 81 (1– cos²θ) =1681+1600 cos²θ – 3280cosθ

⇒ 81 – 81cos²θ = 1681 +1600cos² θ – 3280 cosθ

⇒ 1681cos²θ –3280cos θ +1600 = 0

⇒ (41)² cos² θ – 2(41) (40cos θ) + (40)² = 0

⇒ (41cos θ – 40 )² = 0

cosθ = 40 / 41

Extrà Information :-

$\: \: \: \: \: \: \: \: \: \: \: \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Answer 2

Answer:

9 sin θ + 40 cos θ = 41

⇒ 9 sin θ = 41 – 40 cos θ _(i)

Squaring both sides, we get

⇒ 81sin²θ = 1681+1600 cos²θ – 2(41) (40cos θ)

[∵ (a – b)² = a² + b² –2ab]

⇒ 81 (1– cos²θ) =1681+1600 cos²θ – 3280cosθ

⇒ 81 – 81cos²θ = 1681 +1600cos² θ – 3280 cosθ

⇒ 1681cos²θ –3280cos θ +1600 = 0

⇒ (41)² cos² θ – 2(41) (40cos θ) + (40)² = 0

⇒ (41cos θ – 40 )² = 0

cosθ = 40 / 41