The altitudes of triangle are $12, 15$ and $20$ units. The largest angle in the triangle is:

75^{\circ}

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90^{\circ}

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120^{\circ}

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135^{\circ}

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Solution

The correct option is C $90^{\circ}$
Area of triangle $= \frac{1}{2} b a s e \times h e i g h t$
Let the length of the sides of the triangle be $a, b, c$
Then, $\frac{1}{2} \times 12 \times a = \frac{1}{2} \times 15 \times b = \frac{1}{2} \times 20 \times c$
$12 a = 15 b = 20 c = 60 k$
$a = 5 k, b = 4 k, c = 3 k$
Since, $b^{2} + c^{2} = a^{2}$
By converse of Pythagoras theorem, it is a right angled triangle.
Hence, the largest angle is $90^{\circ}$

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The altitudes of triangle are 12,15 and 20 units. The largest angle in the triangle is:

The correct option is C 90∘Area of triangle =12base×heightLet the length of the sides of the triangle be a,b,cThen, 12×12×a=12×15×b=12×20×c12a=15b=20c=60ka=5k,b=4k,c=3kSince, b2+c2=a2By converse of Pythagoras theorem, it is a right angled triangle. Hence, the largest angle is 90∘

The altitudes of triangle are $12, 15$ and $20$ units. The largest angle in the triangle is: