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2 years ago

Triangles ABC and EDC are similar . find the measure of angle D.

Mathematics

1 answer:

ratelena [41]2 years ago

4 0

Since you are given that the triangle are similar, by definition of similar, we know that corresponding angles are congruent and corresponding sides are proportional.
The ORDER of the similarity statement is paramount.
Triangle ABC is similar to triangle EDC means that:
Angle A = Angle E
Angle B = Angle D
Angle C = Angle C

Angle B = 62 degrees so Angle D = 62 degrees.

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<h3>Given:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

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<h3>Solution:-</h3>

$\\ \sf \implies\frac{7 + 3 \sqrt{5} }{3 + \sqrt{5} } - \frac{7 - 3 \sqrt{5} }{3 - \sqrt{5} } = a + \sqrt{5} b \\$

$\\ \sf \implies\frac{( \: 7 + 3 \sqrt{5} \: \: ( 3 - \sqrt{5}) \: \: - 7 - 3 \sqrt{5} \: \: ( 3 + \sqrt{5}) \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 21 - 7 \sqrt{5} \: + 9 \sqrt{5} - 15) \: \: - ( \: 21 + 7 \sqrt{5} \: - 9 \sqrt{5} + 15)\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{( \: 6 + 2 \sqrt{5} ) \: \: - ( \: 6 - 2 \sqrt{5} )\: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 6 + 2 \sqrt{5} \: \: - \: \: 6 - 2 \sqrt{5} \: }{(3 + \sqrt{5}) \: \:(3 + \sqrt{5}) } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{3 {}^{2} - {\sqrt{5} }^{2} } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 9 - 5 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: 4 \sqrt{5} \: \: }{ \: \: \: \: 4 \: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies\frac{\: \cancel{4 } \sqrt{5} \: \: }{ \: \: \: \: \cancel{4 }\: \: \: } = a + \sqrt{5} \: b\\$

$\\ \sf \implies \: \sqrt{5} = a + \sqrt{5} \: b\\$

we can also write it as ;

$\\ \sf \implies \: 0 + \sqrt{5} = a + \sqrt{5} \: b\\$

★<u> </u><u>Henceforth, the value of a and b are</u> :

→ a = 0

→ b = 1

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2 years ago

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