Answer :

Given, AB = AD

BC = CD

AC which is common to both ∆ADC and ∆ABC.

So, ∠DAC = ∠CAB = 30 as AB = AD and AC which is common.

And, ∠DCA = ∠ACB = 50 as BC = CD and AC which is common.

Now, we know sum of all interior angles of a triangle is 180⁰.

In ∆ADC,

∠a + ∠d + ∠c = 180⁰

30⁰ + ∠d + 50⁰ = 180

80⁰ + ∠d = 180⁰

∠d = 180⁰-80⁰

∠d = 100⁰

In ∆ABC,

∠a + ∠b + ∠c = 180⁰

30⁰ + ∠b + 50⁰ = 180⁰

80⁰ + ∠b = 180⁰

∠b = 180⁰-80⁰

∠b = 100⁰

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<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I

<span lang="EN-USKerala Board Mathematics Part I