The altitude AD of in - ABC is 10 cm- The - A is an obtuse angle and BD -10 cm - CD -10 -3 cm - Find - A-A- 75-B- 95-C- 100-D- 105-

The correct option is D 105° tan B = AD/BD tan B = 10 cm/10 cm tan B = 1 So, ∠B = 45^∘ tan C = AD/DC tan C = 10 cm/10√(3) cm tan C = 1/√(3) ∠C = 30^∘ ∠ A = ...

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Standard IX

Mathematics

SAS Criteria for Congruency

The altitude ...

Question

The altitude AD of in $Δ$ ABC is 10 cm. The $∠$ A is an obtuse angle and BD = 10 cm, CD = 10 $\sqrt{3}$ cm. Find $∠$ A.

75°

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95°

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100°

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105°

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Solution

The correct option is D
105°

tan B = $\frac{A D}{B D}$

tan B = $\frac{10 c m}{10 c m}$

tan B = 1

So, $∠$ B = $45^{\circ}$

tan C = $\frac{A D}{D C}$

tan C = $\frac{10 c m}{10 \sqrt{3} c m}$

tan C = $\frac{1}{\sqrt{3}}$

$∠$ C = $30^{\circ}$

$∠$ A = $180^{\circ}$ - $∠$ B - $∠$ C

= $105^{\circ}$

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The altitude AD of in ΔABC is 10 cm. The ∠A is an obtuse angle and BD = 10 cm, CD = 10√3 cm. Find ∠A.

The correct option is D 105° tan B = ADBD tan B = 10 cm10 cm tan B = 1 So, ∠B = 45∘ tan C = ADDC tan C = 10 cm10√3 cm tan C = 1√3 ∠C = 30∘ ∠ A = 180∘ - ∠ B - ∠ C = 105∘

The altitude AD of in $Δ$ ABC is 10 cm. The $∠$ A is an obtuse angle and BD = 10 cm, CD = 10 $\sqrt{3}$ cm. Find $∠$ A.