Question

In a figure ABC is a right triangle with angle B = 90. AB = 5cm and AC = 13cm Find:- 1. sinC 2. cosC
3. tanA

4. secA

5. Sin^2A - cos^2A

6. sec^C - tan^2C

Answer 1

$\Large{\underline{\underline{According\:to\:the\:Question}}}$

In ∆ABC

${\boxed{\sf\:{Using\;Pythagoras\;theorem}}}$

BC² = AC² - AB²

BC² = (13)² - (5)²

BC² = 169 - 25

BC² = 144

BC = √144

BC = 12 cm

★Now we have :-

AB = 5 cm , BC = 12 cm and AC = 13 cm

${\boxed{\sf\:{Now\;substitute\;the\;values}}}$

$\tt{\rightarrow 1)sinC=\dfrac{AB}{AC}=\dfrac{5}{13}}$

$\tt{\rightarrow 2)cosC=\dfrac{BC}{AC}=\dfrac{12}{13}}$

$\tt{\rightarrow 3)tanA=\dfrac{BC}{AB}=\dfrac{12}{5}}$

$\tt{\rightarrow 4)secA=\dfrac{AC}{AB}=\dfrac{13}{5}}$

5. sin²A - cos²A

$\tt{\rightarrow(\dfrac{BC}{AC})^{2}-(\dfrac{AB}{AC})^{2}}$

$\tt{\rightarrow(\dfrac{12}{13})^{2}-(\dfrac{5}{13})^{2}}$

$\tt{\rightarrow\dfrac{144-25}{169}}$

$\tt{\rightarrow\dfrac{119}{169}}$

6. sec²C - tan²C

$\tt{\rightarrow(\dfrac{AC}{BC})^{2}-(\dfrac{AB}{BC})^{2}}$

$\tt{\rightarrow(\dfrac{13}{12})^{2}-(\dfrac{5}{12})^{2}}$

$\tt{\rightarrow\dfrac{169-25}{144}}$

$\tt{\rightarrow\dfrac{114}{114}}$

= 1

Answer 2

*Refer the attachment for figure.

GIVEN : ABC is a triangle having right angled at B. AB = 5 cm and AC = 13 cm

FIND : 1) sinC

2) cosC

3) tanA

4) secA

5) sin²A - cos²A

6) sec²C - tan²C

Solution :

In $\triangle$ABC

By Pythagoras Theorem

=> H² = P² + B²

=> AC² = AB² + BC²

=> AC² - AB² = BC²

=> BC² = AC² - AB²

=> BC² = (13)² - (5)²

=> BC² = 169 - 25

=> BC² = 144

=> BC = 12 cm

_______________________________

1) sinC = P/H

=> AB/AC = 5/13

2) cosC = B/H

=> BC/AC= 12/13

3) tanA = P/B

=> BC/AB = 12/5

4) secA = H/B

=> AC/AB = 13/5

5) sin²A - cos²A = (BC/AC)² - (AB/AC)²

=> (12/13)² - (5/13)²

=> 144/169 - 25/169

=> (144 - 25)/169

=> 119/169

6) sec²C - tan²C = (AC/BC)² - (AB/BC)²

=> (13/12)² - (5/12)²

=> 169/144 - 25/144

=> (169 - 25)/144

=> 144/144

=> 1

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