Question

Write the value of perpendicular, base and hypotenuse if tan = 1, where is an acute angle in

right angled triangle.​

Answer 1

10,10,14.14

Step-by-step explanation:

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Answer 2

$In \: triangle \: ABC, \: \angle {B} = 90\degree$

$and \: \angle {A} = \theta$

$i )tan \theta = 1 \: (given)$

$\implies \frac{ Side \: opposite \:to \: \theta }{adjacent \: side \:to \: \theta} = 1$

$\implies \frac{BC}{AB} = 1$

$\implies BC = AB \: --(1)$

$In \: right \: triangle \: ABC ,$

$AC^{2} = AB^{2} + BC^{2} \\\blue{ ( By \: Pythagoras \:theorem )}$

$\implies AC^{2} = 1^{2} + 1^{2}$

$\implies AC^{2} = 1+ 1$

$\implies AC^{2} = 2$

$\implies AC= \sqrt{2} \: --(2)$

$\red { Hypotenuse (AC) } \green { = \sqrt{2}}$

$ii ) tan \theta = 1$

$\implies tan \theta = tan 45 \degree$

$\implies \theta = 45 \degree\: --(2)$

$iii) Cos 45 \degree = \frac{AB}{AC}$

$\implies \frac{1}{\sqrt{2}} = \frac{AB}{\sqrt{2}}$

$\implies \frac{\sqrt{2}}{\sqrt{2}} = AB$

$\implies\red{ Base (AB) }\green {= 1}$

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