Question

In triangle ABC, angle A = 90°. lf cosB = 40/41, find sinB and tanB

Answer 1

$\bf{\underline{\red{\bf{Given\::}}}}$

★ In triangle ABC, angle A = 90°

★ $\cos{B} = \dfrac{40}{41}$

$\bf{\underline{\red{\bf{To \: find\::}}}}$

★ sinB = ?

★ tanB = ?

$\bf{\underline{\red{\bf{Solution\::}}}}$

$\implies\sf\cos{B} = \dfrac{40}{41}$ = $\sf \dfrac{B}{H}$

• Base of ∆ :- 40 units

• Hypotenuse of ∆ :- 41 units

✩ Now, Using Pythagoras theorem:-

• $\sf (H)^2 = (B)^2 + (P)^2$

• $\sf (41)^2 = (40)^2 + (P)^2$

• $\sf 1681 = 1600 + (P)^2$

• $\sf (P)^2 = 1681 - 1600$

• $\sf (P)^2 = 81$

• $\sf \sqrt{(P^2)} = \sqrt{81}$

• $\sf \cancel{\sqrt{P^2}}= 81$

• $\sf P = 81 units$

✩ Now,

• Perpendicular of ∆ = 81 units

✩ Therefore,

$\implies\sf\sin{B} = \dfrac{P}{H} = \dfrac{81}{41}$

$\implies\sf\tan{B} = \dfrac{P}{B} = \dfrac{81}{40}$

________________________________

Answer 2

$\red{\bold{\underline{\underline{Given}}}}$

In ∆ ABC , ∠A =90°

$cosB = \frac{40}{41}$

$\orange{\bold{\underline{\underline{To Find:-}}}}$

CosB =?

TanB =?

$\red{\bold{\underline{\underline{Step \:by\: step\:explanation:-}}}}$

$\implies \: cosB = \frac{B}{H} = \frac{40}{41}$

Here,

Base of triangle (∆) = 40unit

Hypotenuse of triangle(∆) = 41 unit

$\color{plum} {\tt {\bigstar}} \ \displaystyle {\tt{Now \: using \: Pythagoras\: theorem\: we \: get, }}$

$\sf \implies \: {H}^{2} = {B}^{2} + {P}^{2}$

$\implies \: {41}^{2} = {40}^{2} + {P}^{2}$

$\implies \: 1681 = 1600 + {P}^{2}$

$\implies \: {P}^{2} = 81$

$\implies \: \sqrt{ {P}^{2} } = \sqrt{81}$

$\implies \: { {P}^{2} } = 81$

$\implies \: P = 81units$

$\pink{\bold{\underline{\underline{Now\: perpendicular \:of\:triangle\: = 81 units}}}}$

$\bf\therefore \: sinB = \frac{P}{H} = \frac{81}{41}$

$\bf\therefore \: tanB = \frac{P}{B} = \frac{81}{40}$

$\green{\tt{\underline{\underline{Be Brainly}}}}$