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Answer 1

✰ Question ༄ :

$In \: triangle \: ABC , \: right \: angled \: at \: B , \\ \: if \: tan \: A = \frac{1}{ \sqrt{3} } , \: find \: the \: value \: of : \\ \\ (i) \: sin \: a \: sin \: c \: + cos \: a \: sin \: c \\ \\ (ii) \: cos \: a \: cos \: c - sin \: a \: sin \: c$

✰ Given ༄ :

(1) Perpendicular = 1

(2) Base = √3 and ,

$(3) \: tan \: a \: = \frac{1}{ \sqrt{3} } = \frac{P }{B} \:$

✰ To find ༄ :

• Hypotenuse of triangle
• sin a cos c + cos a sin c
• cos a cos c - sin a sin c

✰ Solution ༄ :

Finding Hypotenuse by Pythagoras Theorem as the given triangle is right angled triangle :

• H² = P² + B²

• H² = ( 1 )² + ( √3 )²

• H² = 1 + 3

• H² = 4

• H = √4

• H = 2

Now values :

$\implies \: sin \: a \: = \frac{1}{2} \\ \\ \implies \:cos \: c = \frac{1}{2} \\ \\ \implies \:sin \: c = \frac{ \sqrt{3} }{2} \\ \\ \implies \:cos \: a = \frac{ \sqrt{3} }{2}$

So ,

Solving for first part :

$(i) \: sin \: a \: sin \: c \: + cos \: a \: sin \: c \\ \\ \implies \: \frac{1}{2} \times \frac{1}{2} + \frac{ \sqrt{3} }{2} \times \frac{ \sqrt{3} }{2} \\ \\ \implies \: \frac{1}{4} + \frac{3}{4} \\ \\ \implies \: \cancel\frac{4}{4} \\ \\ \implies \: \boxed{ \bold \pink{1}}$

Solving for second part :

$(ii) \: cos \: a \: cos \: c - sin \: a \: sin \: c \\ \\ \implies \: \frac{ \sqrt{3} }{2} \times \frac{1}{2} - \frac{1}{2} \times \frac{ \sqrt{3} }{2} \\ \\ \implies \: \frac{ \sqrt{3} }{4} - \frac{ \sqrt{3} }{4} \\ \\ \implies \: \boxed{ \bold \pink{0}}$

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✰ Question ༄ :

In triangle PQR , right angled at Q , PR + QR = 25 cm and PQ = 5 cm . Determine the values of sin P , cos P and tan P .

✰ Given ༄ :

• PR + QR = 25 cm

• PQ = 5 cm

✰ To find ༄ :

Value of ,

• (1) sin p

• (2) cos p

• (3) tan p

✰ Solution ༄ :

• By Pythagoras Theorem finding QR :

• PR² = PQ² + QR²

In question it is given that , PR + QR = 25 cm , So value of PR = 25 - QR .

Now ,

Value of QR :

• ( 25 - QR )² = ( 5 )² + ( QR )²

• 625 + QR² - ( 2 × 25 × QR ) = 25 + QR²

• 625 - 50 QR = 25 + QR² - QR²

• 625 - 50 QR = 25

• - 50 QR = 25 - 625

• - 50 QR = - 600

• QR = 600/25

• QR = 12 cm

Value of PR :

• PR + QR = 25

• PR + 12 = 25

• PR = 25 - 12

• PR = 13 cm

Therefore ,

$(1) \: sin \: p \: = \frac{P }{</u></strong><strong><u>H</u></strong><strong><u>} \\ \\ \implies \: \boxed{ \bold \pink{ sin \: p = \frac{12}{13} }} \\ \\ (2) \: cos \: p \: = \frac{B}{H} \\ \\ \implies \: \boxed{ \bold \pink{cos \: p \: = \frac{5}{13} }} \\ \\ (3) \: tan \: p = \frac{P}{B} \\ \\ \implies \: \boxed{ \bold\pink{tan \: p \: = \frac{12}{5} }}$

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✰ ADITIONAL ✰

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

Trigonometric Identities :

➱ Cos² θ + Sin² θ = 1.

➱ 1 + Tan² θ = Sec² θ

➱ 1 + Cot² θ = Cosec² θ

Answer 2

Answer:

this is the right answer