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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 114 of 150

Marks: +1, -0

\triangle A B C

a=13, b=14, c=15

\sin A+\cos A

is...

Solution:

Concept:

Use the cosine rule to find

\cos A

from the three sides, then use the Pythagorean identity to find

\sin A

Explanation:

\triangle ABC

, the sides are

a=13

b=14

c=15

Here, side

a

is opposite angle

A

By the cosine rule,

a^{2}=b^{2}+c^{2}-2bc\cos A

So,

\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}

Substitute the given values:

\cos A=\frac{14^{2}+15^{2}-13^{2}}{2 \times 14 \times 15}

\cos A=\frac{196+225-169}{420}=\frac{252}{420}=\frac{3}{5}

Now, using

\sin^{2}A+\cos^{2}A=1

\sin^{2}A=1-\left(\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}

Since angle

A

lies in a triangle,

\sin A

is positive.

\sin A=\frac{4}{5}

Therefore,

\sin A+\cos A=\frac{4}{5}+\frac{3}{5}=\frac{7}{5}

Answer:

The correct option is B, i.e.

\frac{7}{5}

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