Question

3. From the given figure, write down the values of:
(i) Sin B
(ii) tan B
(iii) Cos C
(iv) Cot C
(v) (Sin B Cos C + Cos B Sin C)
(vi) (Sec² C – tan² C)​

Answer 1

In a right angled triangle ∆ABC :

• AC : Perpendicular : ?
• AB : Base : 15 units
• BC : Hypotenuse : 17 units

From Pythagoras theorem, we get :

$:\implies~~{\bf{\purple{(AC)^2+(AB)^2=(BC)^2}}}$

$\sf:\implies~~ (AC)^2+(15)^2=(17)^2$

$\sf:\implies~~ (AC)^2+225=289$

$\sf :\implies~~ (AC)^2=289-225$

$\sf:\implies~~ (AC)^2=64$

$\sf:\implies~~ AC=\sqrt{64}$

$\purple{:\implies~~\bf{ AC=8~units}}$

$~~~~~~~~$_____________

${ \bf{ \purple{(~I~)~Sin B : }}}$

We know that :

$\sf :\implies~~ SinB=\dfrac{Opposite}{Hypotenuse}$

$\sf :\implies~~SinB =\dfrac{AC}{BC}$

$\purple{:\implies~~ \bf SinB =\dfrac{8}{17}}$

$\:$

$~~~~~~~~$_____________

$\purple{\bf{(~II~)~TanB:}}$

We know that :

$\sf :\implies~~ TanB=\dfrac{Opposite}{Adjacent}$

$\sf :\implies~~ TanB=\dfrac{AC}{AB}$

$\purple{:\implies~~ \bf{TanB=\dfrac{8}{15}}}$

$~~~~~~~~$_____________

${\bf{\purple{(~III~)~CosC:}}}$

We know that :

$\sf :\implies~~ CosC=\dfrac{Adjacent}{Hypotenuse}$

$\sf:\implies~~ CosC=\dfrac{AC}{BC}$

$\purple{:\implies~~ {\bf{CosC=\dfrac{8}{17}}}}$

$~~~~~~~~$_____________

${\bf{\purple{(~IV~)~CotC}}}$

We know that :

$\sf:\implies~~ CotC=\dfrac{Adjacent}{Opposite}$

$\sf:\implies~~ CotC=\dfrac{AC}{AB}$

$\purple{:\implies~~{\bf{CotC=\dfrac{8}{15}}}}$

$~~~~~~~~$_____________

${\bf{\purple{(~V~)~SinB~CosC+CosB~SinC}}}$

We already know the value of SinB and CosC. Now let's find the value of :

• CosB
• SinC

We know :

$\sf:\implies~~ CosB=\dfrac{Adjacent}{Hypotenuse}$

$\sf:\implies~~ CosB=\dfrac{AB}{BC}$

$\purple{:\implies~~{\bf{CosB=\dfrac{15}{17}}}}$

Also :

$\sf :\implies~~ SinC=\dfrac{Opposite}{Hypotenuse}$

$\sf:\implies~~ SinC=\dfrac{AB}{BC}$

$\purple{:\implies~~{\bf{ SinC=\dfrac{15}{17}}}}$

Finally, substituting :

$\sf : \implies \: \: SinB~CosC+CosB~SinC=\bigg( \dfrac{8}{17} \times \dfrac{8}{17} \bigg) + \bigg( \dfrac{15}{17} + \dfrac{15}{17} \bigg )$

$\sf : \implies \: \: SinB~CosC+CosB~SinC=\dfrac{64}{289} + \dfrac{225}{289}$

$\sf : \implies \: \:SinB~CosC+SinC~CosB = \dfrac{64 + 225}{289}$

$\sf : \implies \: \: SinB~CosC+SinC~CosB= \dfrac{289}{289}$

$\purple{:\implies~~{\bf{SinB~CosC+CosB~SinC=1}}}$

$~~~~~~~~$_____________

${\bf{\purple{Sec^2C-Tan^2C}}}$

Firstly, let's find the value of SecC :

$\sf:\implies~~SecC=\dfrac{Hypotenuse}{Adjacent}$

$\sf:\implies~~ SecC=\dfrac{BC}{AC}$

$\purple{:\implies~~{\bf{SecC=\dfrac{17}{8}}}}$

Now, let's find TanC :

$\sf:\implies~~ TanC=\dfrac{Opposite}{Adjacent}$

$\sf:\implies~~ TanC=\dfrac{AB}{AC}$

$\purple{:\implies~~{\bf{TanC=\dfrac{15}{8}}}}$

Now, substituting :

$\sf :\implies~~ Sec^2-Tan^2C=\bigg(\dfrac{17}{8}\bigg)^2-\bigg(\dfrac{15}{8}\bigg)^2$

$\sf : \implies \: \: Sec^2C-Tan^2C=\cfrac{289}{64} - \cfrac{225}{64}$

$\sf : \implies \: \: Sec^2C-Tan^2C= \dfrac{289 - 225}{64}$

$\sf : \implies \: \: Sec^2-Tan^2C=\cfrac{64}{64}$

$\purple{:\implies~~{\bf{Sec^2C-Tan^2C=1}}}$