Menu Close

Question-215943

Tinku Tara 0 Comments
Pin




Question Number 215943 by sitholenonto last updated on 22/Jan/25
Answered by MathematicalUser2357 last updated on 23/Jan/25
Q.2.1 (√((30−10×tan 60°)^2 +10^2 )) 16.14836✓ Q.2.2 (√((200−400×cos 70°)^2 +(200×tan 60°+400×sin 70°)^2 )) 725.04623✓ tan^(−1) (((200×tan 60°+400×sin 70°)/(200−400×cos 70°)))/(1°) 85(°)✓ Q.2.3(Following Q.2.2) (((200−400×cos 70°)×20)/(10^3 )) 1.263839✓
$$\mathrm{Q}.\mathrm{2}.\mathrm{1} \\ $$$$\sqrt{\left(\mathrm{30}−\mathrm{10}×\mathrm{tan}\:\mathrm{60}°\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} } \\ $$$$\mathrm{16}.\mathrm{14836}\checkmark \\ $$$$\mathrm{Q}.\mathrm{2}.\mathrm{2} \\ $$$$\sqrt{\left(\mathrm{200}−\mathrm{400}×\mathrm{cos}\:\mathrm{70}°\right)^{\mathrm{2}} +\left(\mathrm{200}×\mathrm{tan}\:\mathrm{60}°+\mathrm{400}×\mathrm{sin}\:\mathrm{70}°\right)^{\mathrm{2}} } \\ $$$$\mathrm{725}.\mathrm{04623}\checkmark \\ $$$$\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{200}×\mathrm{tan}\:\mathrm{60}°+\mathrm{400}×\mathrm{sin}\:\mathrm{70}°}{\mathrm{200}−\mathrm{400}×\mathrm{cos}\:\mathrm{70}°}\right)/\left(\mathrm{1}°\right) \\ $$$$\mathrm{85}\left(°\right)\checkmark \\ $$$$\mathrm{Q}.\mathrm{2}.\mathrm{3}\left(\mathrm{Following}\:\mathrm{Q}.\mathrm{2}.\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{200}−\mathrm{400}×\mathrm{cos}\:\mathrm{70}°\right)×\mathrm{20}}{\mathrm{10}^{\mathrm{3}} } \\ $$$$\mathrm{1}.\mathrm{263839}\checkmark \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *