Question

if ∆abc~∆pqr angle a+angle b=100° then angle r=​

Answer 1

Answer:

$\boxed{\bf \: \angle \: r = {80}^{ \circ} \: }$

Step-by-step explanation:

Given that,

$\sf \: \triangle \: abc \: \sim \: \triangle \: pqr$

So, by corresponding parts of similar triangles, we have

$\implies\sf \: \angle \: a = \angle \: p$

$\implies\sf \: \angle \: b = \angle \: q$

$\implies\sf \: \angle \: c= \angle \: r$

Further given that,

$\sf \: \angle \: a + \angle \: b = {100}^{ \circ}$

can be further rewritten as

$\sf \: \angle \: p + \angle \: q= {100}^{ \circ}$

$\left[ \because \:\sf \: \angle \: a = \angle \: p, \: \: \angle \: b = \angle \: q \right]$

Now, In triangle pqr,

We know, sum of all interior angles of a triangle is 180°.

$\sf \: \angle \: p + \angle \: q + \angle \: r= {180}^{ \circ}$

$\sf \: {100}^{ \circ} + \angle \: r= {180}^{ \circ}$

$\left[ \because \: \sf \: \angle \: p + \angle \: q = {100}^{ \circ}\right]$

$\sf \: \angle \: r= {180}^{ \circ} - {100}^{ \circ}$

$\implies\sf \: \angle \: r = {80}^{ \circ}$

Hence,

$\implies\sf \: \boxed{\bf \: \angle \: r = {80}^{ \circ} \: }$

Answer 2

Step-by-step explanation:

Hope this helps you!!!!!