Math, asked by vishnurajoria, 3 months ago

2) Find the values of

i) 5 sin 30° + 3 tan 45°

ii) 4/5 tan^2 60° + 3 sin^2 60°

iii) 2 sin 30° + cos 0° + 3 sin 90°

iv) tan 60 / (sin 60 + cos 60)

v) cos^2 45° + sin^2 30°

vi) cos 60° × cos 30° + sin 60° × sin 30°

3) If sin theta = 4/5 then find cos theta

4) If cos theta = 15/17 then find sin theta

please fast solve
i give point 5 each​

Answers

Answered by Anonymous
51

2) Find the values of :-

\bf \large \red{{i) \: 5 \: sin \: 30 \degree \:  +  \: 3  \: \tan \: 45 \degree}}\\ \\

\bf \large{ =  \: 5 \times  \dfrac{1}{2}  \: +  \: 3 \times 1 }\\ \\

 \bf  \large{ =  \: \dfrac{5}{2}  \:  +  \:  \dfrac{3}{1}}\\ \\

\bf  \large{ =  \: \dfrac{5 \times 1 \:  +  \: 3 \times 2}{2  \times 1} }\\ \\

\bf  \large{ =  \:  \dfrac{5 \:  + \:  6}{2} }\\ \\

\bf  \large \purple{  =  \: \dfrac{11}{2} }

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\bf  \large \red{ii) \:  \:  \dfrac{4}{5} \:  {tan}^{2}  \: 60 \degree \:  +  \: 3 \:  {sin}^{2}  \: 60 \degree }\\ \\

\bf  \large{  =  \: \dfrac{4}{5} \times  (\sqrt{3})^{2} \:  +  \: 3 \:  \times  \Large{\binom{ \sqrt{3} }{2}}^{2} }\\ \\

\bf \large{  =  \:  \: \dfrac{4}{5}  \times 3 \:  +  \: 3 \times  \dfrac{3}{4} }\\ \\

\bf \large{ =  \:  \:  \dfrac{12}{5} \:  +  \:  \dfrac{9}{4}  }\\ \\

\bf \large{ =  \:  \:  \dfrac{12 \times 4 \:  + \:  9 \times 5}{5 \times 4} }\\ \\

\bf \large{ =  \:  \:  \dfrac{48 \:  + \:  45}{20} }\\ \\

\bf \large \purple{ =  \:  \:  \dfrac{93}{20} }\\ \\

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\bf  \large \red{iii) \:  \: 2 \: sin \: 30 \degree \:  +  \: cos \: 0 \degree \:  +  \: 3 \: sin \: 90 \degree }\\ \\

 \bf\large{ =  \:  \: \cancel 2 \times \dfrac{1}{ \cancel2} \:  + \:  1 \:  +  \: 3  \times 1}\\ \\

 \bf \large{ =  \:  \: 1 \:  +  \: 1 \:  +  \: 3}\\ \\

 \bf \large \purple{ =  \:  \: 5}\\ \\

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 \bf \large \red{ iv) \:  \: \dfrac{tan \: 60}{sin \: 60 \:  +  \: cos \: 60} }\\ \\

 \bf \large{ =  \:  \:  \dfrac{ \sqrt{3} }{ \dfrac{ \sqrt{3} }{2} \:  +  \:  \dfrac{1}{2}  } }\\ \\

 \bf \large{ =  \:  \:  \dfrac{ \sqrt{3} }{ \dfrac{ \sqrt{3}  + 1}{2}}}\\ \\

\bf \large{ =  \:  \:  \dfrac{ \sqrt{3} }{1}  \: \times  \:  \dfrac{2}{ \sqrt{3} \:  +  \: 1 } }\\ \\

\bf \large \purple{ =  \:  \:  \dfrac{2  \: \sqrt{3} }{ \sqrt{3}  \:  + \:  1 }}\\ \\

\bf \large \green{If \: solved \: further \: :- }\\ \\

\bf\large\green{By \: rationalisation \: :}  \\  \\

\bf\large{ = \dfrac{2\sqrt{3}}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}}\\  \\  \\

\bf\large{ =\dfrac{(2\sqrt{3})(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)}} \\  \\

\bf\large{= \dfrac{2(\sqrt{3})^{2} - 2\sqrt{3}}{(\sqrt{3} + 1)(\sqrt{3} - 1)}} \\  \\

\bf\large{=  \dfrac{2 \times 3 - 2\sqrt{3}}{(\sqrt{3} + 1)(\sqrt{3} - 1)}} \\  \\

\bf\large{= \dfrac{6  - 2\sqrt{3}}{(\sqrt{3} + 1)(\sqrt{3} - 1)}}\\  \\

\bf\large\blue{By  \: using \:  identity \:  (a+b)(a-b) \:  =  \:  {a}^{2}  -  {b}^{2} , we  \: have : }\\  \\

\bf\large{ = \dfrac{6  - 2\sqrt{3}}{(\sqrt{3})^{2} - (1)^{2}}} \\  \\

\bf\large {=  \dfrac{6  - 2\sqrt{3}}{3 - 1}} \\  \\

\bf\large{= \dfrac{6  - 2\sqrt{3}}{2}}\\  \\

\bf\large{=\dfrac{\cancel{2}(3  - \sqrt{3})}{\cancel{2}}} \\  \\

 \bf\large\purple{= 3  - \sqrt{3}} \\  \\

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 \bf \large \red{ v) \:  \: {cos}^{2} \: 45 \degree \:  +  \:  {sin}^{2}  \: 30 \degree }\\ \\

 \bf \large{  =  \:  \: \binom{1}{ \sqrt{2} } ^{2}  \:  +   \: \binom{1}{2}^{2}  }\\ \\

 \bf \large{  =  \:  \: \dfrac{1}{2}  \: + \:  \dfrac{1}{4}  }\\ \\

 \bf \large{ =  \:  \:  \dfrac{1 \times 4 \:  +  \: 1 \times 2}{2 \times 4} }\\ \\

 \bf \large{ =  \:  \:  \dfrac{4 \:  +  \: 2}{8} }\\ \\

 \bf \large{ =  \:  \:  \dfrac{6}{8} }\\ \\

 \bf \large \purple{ =  \:  \:  \dfrac{3}{4} }\\ \\

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 \bf \large \red{vi) \:  \: cos \: 60 \degree \:  \times  \: cos \: 30 \degree \:  +  \: sin \: 60 \degree \:  \times sin \: 30 \degree }\\ \\

\bf \large{ =  \:  \: \dfrac{1}{2}   \: \times  \:  \dfrac{ \sqrt{3} }{2}   \: +  \:  \dfrac{ \sqrt{3} }{2}   \: \times  \:  \dfrac{1}{2} }\\ \\

\bf \large{ =  \:  \:  \dfrac{ \sqrt{3} }{4} \:  +  \:   \dfrac{ \sqrt{3} }{4} }\\ \\

\bf \large{  =  \:  \: \dfrac{ \sqrt{3} +  \sqrt{3}  }{4} }\\ \\

\bf \large{  =  \:  \: \dfrac{ \cancel2 \sqrt{3} }{ \cancel4} }\\ \\

\bf \large \purple{ =  \:  \:  \dfrac{ \sqrt{3} }{2} }\\ \\

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\bf \large \red{3) \: \: If \: sin  \: \theta \:  =  \dfrac{4}{5}  \: \: then \: find \: cos \:  \theta }\\ \\

\bf \large \purple{Refer \: to \: the \: 1st \: attachment \: for \: the \: answer \: !}\\ \\

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\bf \large\red{4) \: \: If \: cos \:  \theta \:  =  \dfrac{15}{17} \:  \: then \: find  \: sin \:  \theta}\\ \\

\bf \large \purple{Refer \: to \: the \: 2nd \: attachment \: for \: the \: answer \: !}\\ \\

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Attachments:

assingh: Well explained!
Anonymous: thank you ^^
Answered by prajaypal21
5

Step-by-step explanation:

1) 11/2

2) 93/20

3) 5

4) 3-3√

5) 3/4

6) √3/2

7) ...

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