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Sia ? 6 years, 5 months ago
Consider two right triangles ABC and PQR in which {tex} \angle B{/tex} and {tex}\angle Q{/tex} are the right angles.
We have,

In {tex}\triangle ABC{/tex}
{tex}\sin B=\frac{AC}{AB}{/tex}
and, In {tex}\triangle PQR{/tex}
{tex}\sin Q=\frac{PR}{PQ}{/tex}
{tex} \because \quad \sin B = \sin Q{/tex}
{tex} \Rightarrow \quad \frac { A C } { A B } = \frac { P R } { P Q }{/tex}
{tex} \Rightarrow \quad \frac { A C } { P R } = \frac { A B } { P Q } = k{/tex}(say) ...... (i)
{tex} \Rightarrow {/tex} AC = kPR and AB = kPQ .....(ii)
Using Pythagoras theorem in triangles ABC and PQR, we obtain
AB2 = AC2 + BC2 and PQ2 = PR2 + QR2
{tex} \Rightarrow \quad B C = \sqrt { A B ^ { 2 } - A C ^ { 2 } } \text { and } Q R = \sqrt { P Q ^ { 2 } - P R ^ { 2 } }{/tex}
{tex} \Rightarrow \quad \frac { B C } { Q R } = \frac { \sqrt { A B ^ { 2 } - A C ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = \frac { \sqrt { k ^ { 2 } P Q ^ { 2 } - k ^ { 2 } P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } }{/tex} [ using (ii) ]
{tex} \Rightarrow \quad \frac { B C } { Q R } = \frac { k \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } { \sqrt { P Q ^ { 2 } - P R ^ { 2 } } } = k{/tex}...(iii)
From (i) and (iii), we get
{tex} \frac { A C } { P R } = \frac { A B } { P Q } = \frac { B C } { Q R }{/tex}
{tex} \Rightarrow \quad \Delta A C B - \Delta P R Q{/tex} [By S.A.S similarity]
{tex} \therefore \quad \angle B = \angle Q{/tex}
Hence proved.
Posted by Dev Ji 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the number be (3q + r)
{tex}n = 3 q + r \quad 0 \leq r < 3{/tex}
{tex}\text { or } 3 q , 3 q + 1,3 q + 2{/tex}
{tex}\text { If } n = 3 q \text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}
{tex}3 q \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 1 \text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}
{tex}( 3 q + 3 ) \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 2 \text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}
{tex}( 3 q + 6 ) \text { is divisible by } 3{/tex}.
{tex}\therefore \text { out of } n , ( n + 2 ) \text { and } ( n + 4 ) \text { only one is divisible by } 3{/tex}.
Posted by Dev Ji 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
- Let us find HCF of 396 and 231 using Euclid’s division algorithm
{tex}\begin{array}{l}396=231\times1+165\\231=165\times1+66\\165=66\times2+33\\66=33\times2+0\\So\;HCF(396,231)=33\\\end{array}{/tex}
So 33 is common factor of 396 and 231
and co-prime numbers have common factor of 1 only.
∴ The 396 and 231 are not co-prime. -
Here we have to find out HCF of 2160 and 847 by Using Euclid’s division Lemma, we get
2160 = 847{tex}\times{/tex}2 + 466
Also 847 = 466{tex}\times{/tex}1 + 381
466 = 381{tex}\times{/tex}1 + 85
381 = 85{tex}\times{/tex}4 + 41
85 = 41{tex}\times{/tex}2 + 3
41=3{tex}\times{/tex}13 + 2
3 = 2{tex}\times{/tex}1 + 1
2 = 1{tex}\times{/tex}2 + 0
{tex}\therefore{/tex}HCF = 1.
Hence the numbers are co-prime.
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Kanchana Raja 7 years, 8 months ago
1Thank You